CONCRETE

MATHEMATICS

Dedicated to Leonhard Euler (1707-l 783)

CONCRETE

MATHEMATICS

Dedicated to Leonhard Euler (1707-l 783)

CONCRETE

MATHEMATICS

Ronald L. Graham

AT&T Bell Laboratories

Donald E. Knuth

Stanford University

Oren Patashnik

Stanford University

ADDISON-WESLEY PUBLISHING COMPANY

Reading, Massachusetts

Menlo Park, California

New York

Don Mills, Ontario

Wokingham, England

Amsterdam

Bonn

Sydney Singapore Tokyo Madrid

San Juan

Library of Congress Cataloging-in-Publication Data

Graham, Ronald Lewis,

1935-

Concrete mathematics

: a foundation for computer science

Ron-

ald L. Graham, Donald

Knuth,

Oren

Patashnik.

xiii,625 p.

24 cm.

Bibliography: p. 578

Includes index.

ISBN o-201-14236-8

Mathematics--1961-

2. Electronic data processing--Mathematics.

I. Knuth, Donald Ervin,

1938-

II. Patashnik,

Oren,

1954-

III. Title.

QA39.2.C733

1988

510--dc19

88-3779

CIP

Sixth printing, with corrections, October 1990

retrieval system or transmitted, in any form or by any means, electronic, mechani-

cal,

photocopying, recording, or otherwise, without the prior written permission of

the publisher. Printed in the United States of America. Published simultaneously

in Canada.

FGHIJK-HA-943210

Preface

“A

odience, level,

and treatment

a description of

such matters is

what prefaces are

supposed to be

about.”

P. R. Halmos

11421

“People do

acquire

a little brief author-

ity

equipping

themselves with

jargon:

they can

pontificate

and

air

superficial

expertise.

But

what we should

ask of educated

mathematicians is

not

what

they can

speechify

about,

nor even what they

know

about

the

existing corpus

of mathematical

knowledge, but

rather what

can

they now do with

their learning and

whether

they can

actually solve

math-

ematical

problems

arising in practice.

In short, we look for

deeds not words.”

Hammersley

[145]

THIS BOOK IS BASED on a course of the same name that has been taught

annually at Stanford University since 1970. About fifty students have taken it

each year-juniors and seniors, but mostly graduate students-and alumni

of these classes have begun to spawn similar courses elsewhere. Thus the time

seems ripe to present the material to a wider audience (including sophomores).

It was a dark and stormy decade when Concrete Mathematics was born.

Long-held values were constantly being questioned during those turbulent

years; college campuses were hotbeds of controversy. The college curriculum

itself was challenged, and mathematics did not escape scrutiny. John Ham-

mersley had just written a thought-provoking article “On the enfeeblement of

mathematical skills by ‘Modern Mathematics’ and by similar soft intellectual

trash in schools and universities”

[145];

other worried mathematicians

[272]

even asked, “Can mathematics be saved?” One of the present authors had

embarked on a series of books called The Art of Computer Programming, and

in writing the first volume he (DEK) had found that there were mathematical

tools missing from his repertoire; the mathematics he needed for a thorough,

well-grounded understanding of computer programs was quite different from

what he’d learned as a mathematics major in college. So he introduced a new

course, teaching what he wished somebody had taught him.

The course title “Concrete Mathematics” was originally intended as an

antidote to “Abstract Mathematics,” since concrete classical results were rap-

idly being swept out of the modern mathematical curriculum by a new wave

of abstract ideas popularly called the “New Math!’ Abstract mathematics is a

wonderful subject, and there’s nothing wrong with it: It’s beautiful, general,

and useful. But its adherents had become deluded that the rest of mathemat-

ics was inferior and no longer worthy of attention. The goal of generalization

had become so fashionable that a generation of mathematicians had become

unable to relish beauty in the particular, to enjoy the challenge of solving

quantitative problems, or to appreciate the value of technique. Abstract math-

ematics was becoming inbred and losing touch with reality; mathematical ed-

ucation needed a concrete counterweight in order to restore a healthy balance.

When DEK taught Concrete Mathematics at Stanford for the first time,

he explained the somewhat strange title by saying that it was his attempt

vi PREFACE

to teach a math course that was hard instead of soft. He announced that,

contrary to the expectations of some of his colleagues, he was not going to

teach the Theory of Aggregates, nor Stone’s Embedding Theorem, nor even

the Stone-Tech compactification. (Several students from the civil engineering

department got up and quietly left the room.)

Although Concrete Mathematics began as a reaction against other trends,

the main reasons for its existence were positive instead of negative. And as

the course continued its popular place in the curriculum, its subject matter

“solidified” and proved to be valuable in a variety of new applications. Mean-

while, independent confirmation for the appropriateness of the name came

from another direction, when Z. A. Melzak published two volumes entitled

Companion to Concrete Mathematics

[214].

The material of concrete mathematics may seem at first to be a disparate

bag of tricks, but practice makes it into a disciplined set of tools. Indeed, the

techniques have an underlying unity and a strong appeal for many people.

When another one of the authors (RLG) first taught the course in 1979, the

students had such fun that they decided to hold a class reunion a year later.

But what exactly is Concrete Mathematics? It is a blend of continuous

and

diSCRETE

mathematics. More concretely, it is the controlled manipulation

of mathematical formulas, using a collection of techniques for solving prob-

lems. Once you, the reader, have learned the material in this book, all you

will need is a cool head, a large sheet of paper, and fairly decent handwriting

in order to evaluate horrendous-looking sums, to solve complex recurrence

relations, and to discover subtle patterns in data. You will be so fluent in

algebraic techniques that you will often find it easier to obtain exact results

than to settle for approximate answers that are valid only in a limiting sense.

The major topics treated in this book include sums, recurrences, ele-

mentary number theory, binomial coefficients, generating functions, discrete

probability, and asymptotic methods. The emphasis is on manipulative tech-

nique rather than on existence theorems or combinatorial reasoning; the goal

is for each reader to become as familiar with discrete operations (like the

greatest-integer function and finite summation) as a student of calculus is

familiar with continuous operations (like the absolute-value function and in-

finite integration).

Notice that this list of topics is quite different from what is usually taught

nowadays in undergraduate courses entitled “Discrete Mathematics!’ There-

fore the subject needs a distinctive name, and “Concrete Mathematics” has

proved to be as suitable as any other.

The original textbook for Stanford’s course on concrete mathematics was

the “Mathematical Preliminaries” section in

The

Art of Computer Program-

ming

[173].

But the presentation in those 110 pages is quite terse, so another

author (OP) was inspired to draft a lengthy set of supplementary notes. The

“The heart of math-

ematics consists

of concrete exam-

ples and concrete

problems.

”

-P.

R. Halmos

11411

“lt is downright

sinful to teach the

abstract before the

concrete.

”

-Z. A. Melzak

12141

Concrete Ma the-

matics is a bridge

to abstract mathe-

matics.

“The advanced

reader who skips

parts that appear

too elementary may

miss

than

the less advanced

reader who skips

parts that appear

too complex.

”

-G.

Pdlya

[238]

(We’re not bold

enough

to try

Distinuous Math-

ema tics.)

‘I

a concrete

life preserver

thrown to students

sinking in a sea of

abstraction.”

Gottschalk

Math graffiti:

Kilroy wasn’t Haar.

Free the group.

Nuke the kernel.

Power to the n.

N=l

P=NP.

I have only a

marginal interest

in this subject.

This was the most

enjoyable course

I’ve ever had. But

it might be nice

to summarize the

material as you

go along.

PREFACE vii

present book is an outgrowth of those notes; it is an expansion of, and a more

leisurely introduction to, the material of Mathematical Preliminaries. Some of

the more advanced parts have been omitted; on the other hand, several topics

not found there have been included here so that the story will be complete.

The authors have enjoyed putting this book together because the subject

began to jell and to take on a life of its own before our eyes; this book almost

seemed to write itself. Moreover, the somewhat unconventional approaches

we have adopted in several places have seemed to fit together so well, after

these years of experience, that we can’t help feeling that this book is a kind

of manifesto about our favorite way to do mathematics. So we think the book

has turned out to be a tale of mathematical beauty and surprise, and we hope

that our readers will share at least

of the pleasure we had while writing it.

Since this book was born in a university setting, we have tried to capture

the spirit of a contemporary classroom by adopting an informal style. Some

people think that mathematics is a serious business that must always be cold

and dry; but we think mathematics is fun, and we aren’t ashamed to admit

the fact. Why should a strict boundary line be drawn between work and

play? Concrete mathematics is full of appealing patterns; the manipulations

are not always easy, but the answers can be astonishingly attractive. The

joys and sorrows of mathematical work are reflected explicitly in this book

because they are part of our lives.

Students always know better than their teachers, so we have asked the

first students of this material to contribute their frank opinions, as “grafhti”

in the margins. Some of these marginal markings are merely corny, some

are profound; some of them warn about ambiguities or obscurities, others

are typical comments made by wise guys in the back row; some are positive,

some are negative, some are zero. But they all are real indications of feelings

that should make the text material easier to assimilate. (The inspiration for

such marginal notes comes from a student handbook entitled Approaching

Stanford, where the official university line is counterbalanced by the remarks

of outgoing students. For example, Stanford says, “There are a few things

you cannot miss in this amorphous shape which is Stanford”; the margin

says, “Amorphous . . . what the

h***

does that mean? Typical of the

pseudo-

intellectualism around here.” Stanford: “There is no end to the potential of

a group of students living together.” Grafhto: “Stanford dorms are like zoos

without a keeper.“)

The margins also include direct quotations from famous mathematicians

of past generations, giving the actual words in which they announced some

of their fundamental discoveries. Somehow it seems appropriate to mix the

words of Leibniz, Euler, Gauss, and others with those of the people who

will be continuing the work. Mathematics is an ongoing endeavor for people

everywhere; many strands are being woven into one rich fabric.

viii PREFACE

This book contains more than 500 exercises, divided into six categories:

see:

Warmups are exercises that EVERY READER should try to do when first

Concrete

mathemat-

its

meanS

dri,,inp

reading the material.

Basics are exercises to develop facts that are best learned by trying

one’s own derivation rather than by reading somebody else’s,

Homework exercises are problems intended to deepen an understand-

ing of material in the current chapter.

Exam problems typically involve ideas from two or more chapters si-

multaneously; they are generally intended for use in take-home exams

(not for in-class exams under time pressure).

Bonus problems

go beyond what an average student of concrete math-

ematics is expected to handle while taking a course based on this book;

they extend the text in interesting ways.

The homework was

tough but

learned

a lot. It was worth

every hour.

Take-home exams

are vital-keep

them.

Exams were harder

than the homework

led me to exoect.

Research problems

may or may not be humanly solvable, but the ones

presented here seem to be worth a try (without time pressure).

Answers to all the exercises appear in Appendix A, often with additional infor-

mation about related results. (Of course, the “answers” to research problems

are incomplete; but even in these cases, partial results or hints are given that

might prove to be helpful.) Readers are encouraged to look at the answers,

especially the answers to the warmup problems, but only AFTER making a

serious attempt to solve the problem without peeking.

We have tried in Appendix C to give proper credit to the sources of

each exercise, since a great deal of creativity and/or luck often goes into

the design of an instructive problem.

Mathematicians have unfortunately

developed a tradition of borrowing exercises without any acknowledgment;

we believe that the opposite tradition, practiced for example by books and

magazines about chess (where names, dates, and locations of original chess

problems are routinely specified) is far superior. However, we have not been

able to pin down the sources of many problems that have become part of the

folklore. If any reader knows the origin of an exercise for which our citation

is missing or inaccurate, we would be glad to learn the details so that we can

correct the omission in subsequent editions of this book.

The typeface used for mathematics throughout this book is a new design

by Hermann Zapf

[310],

commissioned by the American Mathematical Society

and developed with the help of a committee that included B. Beeton, R. P.

Boas, L. K. Durst, D. E. Knuth, P. Murdock, R. S. Palais, P. Renz, E. Swanson,

S. B. Whidden, and W. B. Woolf. The underlying philosophy of Zapf’s design

is to capture the flavor of mathematics as it might be written by a mathemati-

cian with excellent handwriting. A handwritten rather than mechanical style

is appropriate because people generally create mathematics with pen, pencil,

Cheaters may pass

this course by just

copying the an-

swers, but they’re

only cheating

themselves.

Difficult exams

don’t take into ac-

count students who

have other classes

to prepare for.

I’m unaccustomed

to this face.

Dear

prof:

Thanks

for (1) the

puns,

(2) the subject

matter.

don’t see

how

what I’ve learned

will

ever

help me.

bad a lot of

trou-

ble in this class, but

know it

sharpened

my math

skills

and

thinking

skills.

would

advise

the

casual student to

stay

away

from this

course.

PREFACE ix

or chalk. (For example, one of the trademarks of the new design is the symbol

for zero, ‘0’, which is slightly pointed at the top because a handwritten zero

rarely closes together smoothly when the curve returns to its starting point.)

The letters are upright, not italic, so that subscripts, superscripts, and ac-

cents are more easily fitted with ordinary symbols. This new type family has

been named AM.9 Euler, after the great Swiss mathematician Leonhard Euler

(1707-1783) who discovered so much of mathematics as we know it today.

The alphabets include Euler Text (Aa Bb Cc through Xx Yy Zz), Euler Frak-

tur

(%a23236

cc through Q’$lu

3,3),

and Euler Script Capitals (A’B e through

y Z), as well as Euler Greek

(AOL

ry through

XXY’J,

nw) and special

symbols such as p and K. We are especially pleased to be able to inaugurate

the Euler family of typefaces in this book, because Leonhard Euler’s spirit

truly lives on every page: Concrete mathematics is Eulerian mathematics.

The authors are extremely grateful to Andrei Broder, Ernst Mayr, An-

drew Yao, and Frances Yao, who contributed greatly to this book during the

years that they taught Concrete Mathematics at Stanford. Furthermore we

offer 1024 thanks to the teaching assistants who creatively transcribed what

took place in class each year and who helped to design the examination ques-

tions; their names are listed in Appendix C. This book, which is essentially

a compendium of sixteen years’ worth of lecture notes, would have been im-

possible without their first-rate work.

Many other people have helped to make this book a reality. For example,

we wish to commend the students at Brown, Columbia, CUNY, Princeton,

Rice, and Stanford who contributed the choice graffiti and helped to debug

our first drafts. Our contacts at Addison-Wesley were especially efficient

and helpful; in particular, we wish to thank our publisher (Peter Gordon),

production supervisor (Bette Aaronson), designer (Roy Brown), and copy ed-

itor (Lyn Dupre). The National Science Foundation and the Office of Naval

Research have given invaluable support. Cheryl Graham was tremendously

helpful as we prepared the index. And above all, we wish to thank our wives

(Fan, Jill, and Amy) for their patience, support, encouragement, and ideas.

We have tried to produce a perfect book, but we are imperfect authors.

Therefore we solicit help in correcting any mistakes that we’ve made. A re-

ward of $2.56 will gratefully be paid to the first finder of any error, whether

it is mathematical, historical, or typographical.

Murray Hill, New Jersey

-RLG

and Stanford, California

DEK

May 1988

A Note on Notation

SOME OF THE SYMBOLISM in this book has not (yet?) become standard.

Here is a list of notations that might be unfamiliar to readers who have learned

similar material from other books, together with the page numbers where

these notations are explained:

Notation

lnx

log x

1x1

xmody

{xl

f(x)

XI1

iRz

H’X’

f'"'(z)

Name

natural logarithm: log, x

binary logarithm: log, x

common logarithm: log,

floor: max{n 1 n < x, integer n}

ceiling: min{ n 1 n 3 x, integer n}

remainder: x

y lx/y]

fractional part: x mod 1

indefinite summation

Page

262

435

definite summation

falling factorial power: x!/(x

n)!

rising factorial power:

T(x

n)/(x)

subfactorial:

n!/O!

n!/l ! + . . + (-1 )“n!/n!

real part: x, if

= x + iy

imaginary part: y, if

= x +

harmonic number: 1

. . . + 1

generalized harmonic number: 1

/lx

. . . + 1

/nx

mth derivative of f at

194

263

456

If you don’t under-

stand what the

x denotes at the

bottom of this page,

try asking your

Latin professor

instead of your

math professor.

[

n-l

Prestressed concrete

mathematics is con-

Crete

mathematics

that’s preceded by

(‘h...%)b

a bewildering list

of notations.

K(al,.

. .

,a,)

iz”l

f(z)

la..@1

[m=nl

[m\nl

Im\nl

[m-l-n1

A NOTE ON NOTATION xi

Stirling cycle number (the “first kind”) 245

Stirling subset number (the “second kind”) 244

Eulerian number 253

Second-order Eulerian number

256

radix notation for

z,“=,

akbk

continuant polynomial 288

hypergeometric function

205

cardinality: number of elements in the set A 39

coefficient of zn in f

(2)

197

closed interval: the set {x 1 016 x 6

(3}

1 if m = n, otherwise 0 * 24

1 if m divides n, otherwise 0 *

102

1 if m exactly divides n, otherwise 0 *

146

1 if m is relatively prime to n, otherwise 0 *

115

*In general, if S is any statement that can be true or false, the bracketed

notation

[S]

stands for 1 if S is true, 0 otherwise.

Throughout this text, we use single-quote marks

(‘.

. .

‘)

to delimit text as

it is written, double-quote marks (“. .

“)

for a phrase as it is spoken. Thus,

Also ‘nonstring’ is

the string of letters ‘string’ is sometimes called a “string!’

a string.

An expression of the form ‘a/be’ means the same as ‘a/(bc)‘. Moreover,

logx/logy = (logx)/(logy) and 2n! = 2(n!).

Contents

Recurrent Problems

1.1 The Tower of Hanoi

1.2 Lines in the Plane 4

1.3 The

Josephus

Problem

Exercises 17

2 Sums

2.1 Notation 21

2.2 Sums and Recurrences 25

2.3 Manipulation of Sums 30

2.4 Multiple Sums 34

2.5 General Methods 41

2.6 Finite and Infinite Calculus

2.7 Infinite Sums 56

Exercises 62

3 Integer Functions

3.1 Floors and Ceilings 67

3.2 Floor/Ceiling Applications 70

3.3 Floor/Ceiling Recurrences 78

3.4

‘mod’:

The Binary Operation

3.5 Floor/Ceiling Sums 86

Exercises 95

4 Number Theory 102

4.1 Divisibility 102

4.2 Primes 105

4.3 Prime Examples 107

4.4 Factorial Factors 111

4.5 Relative Primality 115

4.6 ‘mod’: The Congruence Relation

123

4.7 Independent Residues 126

4.8 Additional Applications 129

4.9 Phi and Mu 133

Exercises 144

5 Binomial Coefficients

153

5.1 Basic Identities 153

5.2 Basic Practice 172

xii

CONTENTS xiii

5.3 Tricks of the Trade 186

5.4 Generating Functions 196

5.5 Hypergeometric Functions 204

5.6 Hypergeometric Transformations 216

5.7 Partial Hypergeometric Sums 223

Exercises 230

6 Special Numbers

243

6.1 Stirling Numbers 243

6.2 Eulerian Numbers 253

6.3 Harmonic Numbers 258

6.4 Harmonic Summation 265

6.5 Bernoulli Numbers 269

6.6 Fibonacci Numbers 276

6.7 Continuants 287

Exercises 295

7 Generating Functions

306

7.1 Domino Theory and Change 306

7.2 Basic Maneuvers 317

7.3 Solving Recurrences 323

7.4 Special Generating Functions 336

7.5 Convolutions 339

7.6 Exponential Generating Functions 350

7.7 Dirichlet Generating Functions 356

Exercises 357

8 Discrete Probability

367

8.1 Definitions 367

8.2 Mean and Variance 373

8.3 Probability Generating Functions 380

8.4 Flipping Coins 387

8.5 Hashing 397

Exercises 413

9 Asymptotics

425

9.1 A Hierarchy 426

9.2 0 Notation 429

9.3 0 Manipulation 436

9.4 Two Asymptotic Tricks 449

9.5 Euler’s Summation Formula 455

9.6 Final Summations 462

Exercises 475

A Answers to Exercises

483

B Bibliography

578

C Credits for Exercises

601

Index 606

List of Tables

624

Recurrent Problems

THIS CHAPTER EXPLORES three sample problems that give a feel for

what’s to come. They have two traits in common: They’ve all been investi-

gated repeatedly by mathematicians; and their solutions all use the idea of

recuvexe,

in which the solution to each problem depends on the solutions

to smaller instances of the same problem.

Raise your hand

if you’ve never

seen this.

OK, the rest of

you can cut to

equation (1.1).

1.1

THE TOWER OF HANOI

Let’s look first at a neat little puzzle called the Tower of Hanoi,

invented by the French mathematician Edouard Lucas in 1883. We are given

a tower of eight disks, initially stacked in decreasing size on one of three pegs:

The objective is to transfer the entire tower to one of the other pegs, moving

only one disk at a time and never moving a larger one onto a smaller.

Lucas

[208]

furnished his toy with a romantic legend about a much larger

Gold -wow.

Tower of Brahma, which supposedly has 64 disks of pure gold resting on three

Are our disks made

of concrete?

diamond needles. At the beginning of time, he said, God placed these golden

disks on the first needle and ordained that a group of priests should transfer

them to the third, according to the rules above. The priests reportedly work

day and night at their task. When they finish, the Tower will crumble and

the world will end.

2 RECURRENT PROBLEMS

It’s not immediately obvious that the puzzle has a solution, but a little

thought (or having seen the problem before) convinces us that it does. Now

the question arises: What’s the best we can do? That is, how many moves

are necessary and sufficient to perform the task?

The best way to tackle a question like this is to generalize it a bit. The

Tower of Brahma has 64 disks and the Tower of Hanoi has 8; let’s consider

what happens if there are n disks.

One advantage of this generalization is that we can scale the problem

down even more. In fact, we’ll see repeatedly in this book that it’s advanta-

geous to LOOK AT SMALL CASES first. It’s easy to see how to transfer a tower

that contains only one or two disks. And a small amount of experimentation

shows how to transfer a tower of three.

The next step in solving the problem is to introduce appropriate notation:

NAME AND CONQUER. Let’s say that

T,,

is the minimum number of moves

that will transfer n disks from one peg to another under Lucas’s rules. Then

is obviously

and

= 3.

We can also get another piece of data for free, by considering the smallest

case of all: Clearly

= 0, because no moves at all are needed to transfer a

tower of n = 0 disks! Smart mathematicians are not ashamed to think small,

because general patterns are easier to perceive when the extreme cases are

well understood (even when they are trivial).

But now let’s change our perspective and try to think big; how can we

transfer a large tower? Experiments with three disks show that the winning

idea is to transfer the top two disks to the middle peg, then move the third,

then bring the other two onto it. This gives us a clue for transferring n disks

in general: We first transfer the n

1 smallest to a different peg (requiring

T,-l

moves), then move the largest (requiring one move), and finally transfer

the n- 1 smallest back onto the largest (requiring another

Tn..1

moves). Thus

we can transfer n disks (for n > 0) in at most

2T,-,

+ 1 moves:

2Tn-1

+ 1 ,

for n > 0.

This formula uses

‘

’

instead of

‘

’

because our construction proves only

that 2T+1 + 1 moves suffice; we haven’t shown that 2T,_, + 1 moves are

necessary. A clever person might be able to think of a shortcut.

But is there a better way? Actually no. At some point we must move the

largest disk. When we do, the n

1 smallest must be on a single peg, and it

has taken at least T,_, moves to put them there. We might move the largest

disk more than once, if we’re not too alert. But after moving the largest disk

for the last time, we must transfer the n- 1 smallest disks (which must again

be on a single peg) back onto the largest; this too requires

T,-

1 moves. Hence

Most of the pub-

lished “solutions”

to Lucas’s problem,

like the early one

of Allardice and

Fraser

[?I,

fail to ex-

plain why

T,,

must

be 3

2T,,

1 + 1.

2Tn-1

+ 1 ,

for n > 0.

Yeah, yeah.

lseen

that

word

before.

Mathematical in-

duction proves that

we can climb as

high as we like on

ladder, by proving

that we can

climb

onto the bottom

rung (the basis)

and that from each

rung we can climb

up to the next one

(the induction).

1.1 THE TOWER OF HANOI 3

These two inequalities, together with the trivial solution for n = 0, yield

=O;

2T+1

for n > 0.

(1.1)

(Notice that these formulas are consistent with the known values

= 1 and

= 3. Our experience with small cases has not only helped us to discover

a general formula, it has also provided a convenient way to check that we

haven’t made a foolish error. Such checks will be especially valuable when we

get into more complicated maneuvers in later chapters.)

A set of equalities like (1.1) is called a recurrence (a.k.a. recurrence

relation or recursion relation). It gives a boundary value and an equation for

the general value in terms of earlier ones. Sometimes we refer to the general

equation alone as a recurrence, although technically it needs a boundary value

to be complete.

The recurrence allows us to compute

T,,

for any n we like. But nobody

really likes to compute from a recurrence, when n is large; it takes too long.

The recurrence only gives indirect,

“local” information. A solution to the

recurrence would make us much happier. That is, we’d like a nice, neat,

“closed form” for

T,,

that lets us compute it quickly, even for large n. With

a closed form, we can understand what

T,,

really is.

So how do we solve a recurrence? One way is to guess the correct solution,

then to prove that our guess is correct. And our best hope for guessing

the solution is to look (again) at small cases. So we compute, successively,

T~=2~3+1=7;T~=2~7+1=15;T~=2~15+1=31;T~=2~31+1=63.

Aha! It certainly looks as if

2n-1,

for n 3 0.

(1.2)

At least this works for n < 6.

Mathematical induction is a general way to prove that some statement

about the integer n is true for all n 3 no. First we prove the statement

when n has its smallest value, no; this is called the basis. Then we prove the

statement for n > no, assuming that it has already been proved for all values

between no and n

inclusive; this is called the induction. Such a proof

gives infinitely many results with only a finite amount of work.

Recurrences are ideally set up for mathematical induction. In our case,

for example, (1.2) follows easily from (1.1): The basis is trivial, since

2’

1 = 0. And the induction follows for n > 0 if we assume that (1.2) holds

when n is replaced by n

T,,

2T,,

2(2

-l)+l

2n-l.

Hence (1.2) holds for n as well. Good! Our quest for

T,,

has ended successfully.

4 RECURRENT PROBLEMS

Of course the priests’ task hasn’t ended; they’re still dutifully moving

disks, and will be for a while, because for n = 64 there are

264-l

moves (about

18 quintillion). Even at the impossible rate of one move per microsecond, they

will need more than 5000 centuries to transfer the Tower of Brahma. Lucas’s

original puzzle is a bit more practical, It requires

1 = 255 moves, which

takes about four minutes for the quick of hand.

The Tower of Hanoi recurrence is typical of many that arise in applica-

tions of all kinds. In finding a closed-form expression for some quantity of

interest like

T,,

we go through three stages:

Look at small cases. This gives us insight into the problem and helps us

in stages 2 and 3.

Find and prove a mathematical expression for the quantity of interest.

What is a proof?

For the Tower of Hanoi, this is the recurrence (1.1) that allows us, given

“One

ha’fofone

the inclination, to compute

T,,

for any n.

percent

pure alco-

hol.

”

Find and prove a closed form for our mathematical expression. For the

Tower of Hanoi, this is the recurrence solution

(1.2).

The third stage is the one we will concentrate on throughout this book. In

fact, we’ll frequently skip stages 1 and 2 entirely, because a mathematical

expression will be given to us as a starting point. But even then, we’ll be

getting into subproblems whose solutions will take us through all three stages.

Our analysis of the Tower of Hanoi led to the correct answer, but it

required an “inductive leap”;

we relied on a lucky guess about the answer.

One of the main objectives of this book is to explain how a person can solve

recurrences without being clairvoyant. For example, we’ll see that recurrence

(1.1) can be simplified by adding 1 to both sides of the equations:

+ 1 = 1;

Lsl

=2T,-,

+2,

for n > 0.

Now if we let

T,,

we have

uo = 1

;

2&-l,

for n > 0.

Interesting: We get

rid of the

(1.1) by adding, not

(1.3) by subtracting.

It doesn’t take genius to discover that the solution to this recurrence is just

= 2”; hence T, = 2”

1. Even a computer could discover this.

1.2 LINES IN THE PLANE

Our second sample problem has a more geometric flavor: How many

slices of pizza can a person obtain by making n straight cuts with a pizza

knife? Or, more academically: What is the maximum number L, of regions

1.2 LINES IN THE PLANE 5

(A pizza with Swiss

cheese?)

A region is convex

if it includes all

line segments

be-

tween any two of its

points. (That’s not

what my dictionary

says, but it’s what

mathematicians

believe.)

defined by n lines in the plane? This problem was first solved in 1826, by the

Swiss mathematician Jacob Steiner

[278].

Again we start by looking at small cases, remembering to begin with the

smallest of all. The plane with no lines has one region; with one line it has

two regions; and with two lines it has four regions:

(Each line extends infinitely in both directions.)

Sure, we think,

= 2”; of course! Adding a new line simply doubles

the number of regions. Unfortunately this is wrong. We could achieve the

doubling if the nth line would split each old region in two; certainly it can

split an old region in at most two pieces, since each old region is convex. (A

straight line can split a convex region into at most two new regions, which

will also be convex.) But when we add the third line-the thick one in the

diagram below- we soon find that it can split at most three of the old regions,

no matter how we’ve placed the first two lines:

Thus

= 4 + 3 = 7 is the best we can do.

And after some thought we realize the appropriate generalization. The

nth line (for n > 0) increases the number of regions by k if and only if it

splits k of the old regions, and it splits k old regions if and only if it hits the

previous lines in k- 1 different places. Two lines can intersect in at most one

point. Therefore the new line can intersect the n- 1 old lines in at most n- 1

different points, and we must have k 6 n. We have established the upper

bound

L-1

+n,

for n > 0.

Furthermore it’s easy to show by induction that we can achieve equality in

this formula. We simply place the nth line in such a way that it’s not parallel

to any of the others (hence it intersects them all), and such that it doesn’t go

6 RECURRENT PROBLEMS

through any of the existing intersection points (hence it intersects them all

in different places). The recurrence is therefore

= 1;

L, = L,-l

+n,

for n > 0.

(1.4)

The known values of

Lz,

and

check perfectly here, so we’ll buy this.

Now we need a closed-form solution. We could play the guessing game

again, but

2, 4, 7,

11,

16, . . . doesn’t look familiar; so let’s try another

tack. We can often understand a recurrence by “unfolding” or “unwinding”

it all the way to the end, as follows:

L, = L,_j + n

L,-z+(n-l)+n

LnP3

+ (n

2) + (n

1) + n

Unfolding?

I’d call this

“plugging in.”

= Lo+1

+2+...

+ (n

2) + (n

1) + n

s,,

where S, = 1 + 2 + 3 + . . + (n

1) + n.

In other words, L, is one more than the sum S, of the first n positive integers.

The quantity S, pops up now and again, so it’s worth making a table of

small values. Then we might recognize such numbers more easily when we

see them the next time:

2 3

5 6 7 8 9 10

12 13 14

3 6 10 15 21

28 36 45 55 66 78

91 105

These values are also called the triangular numbers, because S, is the number

of bowling pins in an n-row triangular array. For example, the usual four-row

array

‘*:::*’

has

pins.

To evaluate S, we can use a trick that Gauss reportedly came up with

in 1786, when he was nine years old

[73]

(see also Euler

[92,

part 1,

$4151):

It seems

a lot

stuff is attributed

s,=

1 + 2 + 3

+...+ (n-l) + n

to Gauss-

either he was really

+Sn=

+ (n-l) + (n-2) + ... + 2 + 1

smart or he had

2S, =

(n+l)

+...+

(n+1)

(n+l)

great press agent.

Maybe

just

We merely add S, to its reversal, so that each of the n columns on the right

sums to n +

Simplifying,

~~~s~n~,!~etic

n(n+l)

’

for n

(1.5)

Actually Gauss is

often called the

greatest mathe-

matician of all time.

So it’s nice to be

able to

understand

at least one of his

discoveries.

When in

doubt,

look at the words.

Why is it Vlosed,”

as opposed to

L’open”?

What

image does it bring

to mind?

Answer: The

equa-

tion is

“closed

”

not

defined

ter;s

itself-not leading

to recurrence.

The

case is “closed” -it

won’t happen again.

Metaphors

are the

key.

“zig” a technical

term?

1.2 LINES IN THE PLANE 7

OK, we have our solution:

n(n+‘)

)

for n 3 0.

(1.6)

As experts, we might be satisfied with this derivation and consider it

a proof, even though we waved our hands a bit when doing the unfolding

and reflecting. But students of mathematics should be able to meet stricter

standards; so it’s a good idea to construct a rigorous proof by induction. The

key induction step is

L, =

L,-lfn

(t(n-l)n+l)+n

tn(n+l)+l.

Now there can be no doubt about the,closed form (1.6).

Incidentally we’ve been talking about “closed forms” without explic-

itly saying what we mean. Usually it’s pretty clear. Recurrences like (1.1)

and (1.4) are not in closed form- they express a quantity in terms of itself;

but solutions like

(1.2)

and (1.6) are. Sums like 1 + 2 + . . . + n are not in

closed form- they cheat by using

’

. . .

‘;

but expressions like n(n + 1)/2 are.

We could give a rough definition like this: An expression for a quantity f(n)

is in closed form if we can compute it using at most a fixed number of “well

known” standard operations, independent of n. For example, 2”

1 and

n(n + 1)/2 are closed forms because they involve only addition, subtraction,

multiplication, division, and exponentiation, in explicit ways.

The total number of simple closed forms is limited, and there are recur-

rences that don’t have simple closed forms. When such recurrences turn out

to be important, because they arise repeatedly, we add new operations to our

repertoire; this can greatly extend the range of problems solvable in “simple”

closed form. For example, the product of the first n integers, n!, has proved

to be so important that we now consider it a basic operation. The formula

‘n!’

is therefore in closed form, although its equivalent ‘1

.2..

. . .n’ is not.

And now, briefly, a variation of the lines-in-the-plane problem: Suppose

that instead of straight lines we use bent lines, each containing one “zig!’

What is the maximum number

of regions determined by n such bent lines

in the plane? We might expect

to be about twice as big as L,, or maybe

three times as big. Let’s see:

8 RECURRENT PROBLEMS

From these small cases, and after a little thought, we realize that a bent

. . and a little

line is like two straight lines except that regions merge when the “two” lines

afterthought...

don’t extend past their intersection point.

’

.:::

. .

(=:

Regions 2, 3, and 4, which would be distinct with two lines, become a single

region when there’s a bent line; we lose two regions. However, if we arrange

things properly-the zig point must lie “beyond” the intersections with the

other lines-that’s all we lose; that is, we lose only two regions per line. Thus

Exercise 18 has the

details.

Lz,-2n

2n(2n+1)/2+1-2n

2n2-n+l,

for

(1.7)

Comparing the closed forms (1.6) and (1.7), we find that for large n,

in’,

2n2;

so we get about four times as many regions with bent lines as with straight

lines. (In later chapters we’ll be discussing how to analyze the approximate

behavior of integer functions when n is large.)

1.3 THE JOSEPHUS PROBLEM

Our final introductory example is a variant of an ancient problem

named for Flavius Josephus, a famous historian of the first century. Legend

has it that Josephus wouldn’t have lived to become famous without his math-

ematical talents. During the Jewish-Roman war, he was among a band of 41

Jewish rebels trapped in a cave by the Romans. Preferring suicide to capture,

the rebels decided to form a circle and, proceeding around it, to kill every

third remaining person until no one was left. But Josephus, along with an

unindicted co-conspirator, wanted none of this suicide nonsense; so he quickly

calculated where he and his friend should stand in the vicious circle.

In our variation, we start with n people numbered 1 to n around a circle,

and we eliminate every second remaining person until only one survives. For

(Ahrens

15,

vol.

and

Herstein

and Kaplansky

11561

discuss the interest-

ing history of this

problem.

Josephus

himself [ISS] is a bit

vague.)

thereby saving

his tale for us to

hear.

1.3 THE JOSEPHUS PROBLEM 9

Here’s a case where

n = 0 makes no

sense.

Even so, a bad

guess isn’t a waste

of time, because it

gets us involved in

the problem.

This is the tricky

part: We have

J(2n)

newnumber(J(n)),

where

newnumber( k) =

2k-1.

example, here’s the starting configuration for n = 10:

9 3

The elimination order is 2, 4, 6, 8, 10, 3, 7, 1, 9, so 5 survives. The problem:

Determine the survivor’s number, J(n).

We just saw that J(l0) = 5. We might conjecture that J(n) = n/2 when

n is even; and the case n = 2 supports the conjecture: J(2) = 1. But a few

other small cases dissuade us-the conjecture fails for n = 4 and n = 6.

n 123456

J(n) 1 1 3 1 3 5

It’s back to the drawing board; let’s try to make a better guess. Hmmm . . .

J(n) always seems to be odd. And in fact, there’s a good reason for this: The

first trip around the circle eliminates all the even numbers. Furthermore, if

n itself is an even number, we arrive at a situation similar to what we began

with, except that there are only half as many people, and their numbers have

changed.

So let’s suppose that we have 2n people originally. After the first

go-

round, we’re left with

2n-1

2n-3

and 3 will be the next to go. This is just like starting out with n people, except

that each person’s number has been doubled and decreased by

That is,

JVn)

2J(n)

for n 3 1

We can now go quickly to large n. For example, we know that J( 10) = 5, so

J(20) = 2J(lO)

1 =

2.5-

1 = 9

Similarly J(40) = 17, and we can deduce that J(5.2”‘) = 2m+’ + 1

10 RECURRENT PROBLEMS

But what about the odd case? With 2n + 1 people, it turns out that

Odd case? Hey,

person number 1 is wiped out just after person number 2n, and we’re left with

leave

brother

out of it.

2n+l

2n-1

Again we almost have the original situation with n people, but this time their

numbers are doubled and increased by 1. Thus

J(2n-t 1) = 2J(n) + 1 ,

for n > 1.

Combining these equations with J( 1) = 1 gives us a recurrence that defines J

in all cases:

J(1)

;

J(2n) =

2J(n)

1 ,

for n >

(1.8)

J(2n + 1) = 2J(n) +

for n 3

Instead of getting J(n) from J(n- l), this recurrence is much more “efficient,”

because it reduces n by a factor of 2 or more each time it’s applied. We could

compute J(

lOOOOOO),

say, with only 19 applications of (1.8). But still, we seek

a closed form, because that will be even quicker and more informative. After

all, this is a matter of life or death.

Our recurrence makes it possible to build a table of small values very

quickly. Perhaps we’ll be able to spot a pattern and guess the answer.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

J(n) 1 1 3 1 3 5 7 1 3 5 7 9 11 13 15 1

Voild!

It seems we can group by powers of 2 (marked by vertical lines in

the table); J(n is always 1 at the beginning of a group and it increases by 2

)

within a group. So if we write n in the form n = 2” + 1, where 2m is the

largest power of 2 not exceeding n and where

is what’s left, the solution to

our recurrence seems to be

J(2” +

2Lf

for m 3 0 and 0 6 1<

2m.

(1.9)

(Notice that if 2” 6 n

< 2 mt’ , the remainder

= n

2” satisfies

2m+’

I”.)

We must now prove (1.9). As in the past we use induction, but this time

the induction is on m. When m = 0 we must have

= 0; thus the basis of

But there’s a sim-

pler way! The

key fact is that

J(2”)

= 1 for

all

and this

follows immedi-

ately from our first

equation,

J(2n)

2J(n)-1.

Hence we know that

the first person will

survive whenever

n isapowerof2.

And in the gen-

eral case, when

n =

2”+1,

the number of

people is reduced

to a power of 2

after there have

been

executions.

The first remaining

person at this point,

the survivor, is

number

21+

1 .

1.3 THE JOSEPHUS PROBLEM

(1.9) reduces to J(1) =

which is true. The induction step has two parts,

depending on whether

is even or odd. If m > 0 and

2”’

2n, then

even and

J(2”

2J(2”-’

l/2)

1 = 2(21/2 + 1)

1 =

21f

1 ,

by (1.8) and the induction hypothesis; this is exactly what we want. A similar

proof works in the odd case, when 2” +

2n + 1. We might also note that

(1.8) implies the relation

J(2nf

J(2n) = 2.

Either way, the induction is complete and (1.9) is established.

To illustrate solution (l.g), let’s compute J( 100). In this case we have

100 =

+ 36, so J(100) = 2.36 + 1 = 73.

Now that we’ve done the hard stuff (solved the problem) we seek the

soft: Every solution to a problem can be generalized so that it applies to a

wider class of problems. Once we’ve learned a technique, it’s instructive to

look at it closely and see how far we can go with it. Hence, for the rest of this

section, we will examine the solution (1.9) and explore some generalizations

of the recurrence (1.8). These explorations will uncover the structure that

underlies all such problems.

Powers of 2 played an important role in our finding the solution, so it’s

natural to look at the radix 2 representations of n and J(n). Suppose n’s

binary expansion is

n =

(b,

b,-l

. .

bo)z

;

that is,

n =

b,2”

+ bmP12mP’ +

...

+ b12 +

bo,

where each

is either 0 or 1 and where the leading bit

is 1. Recalling

that n = 2” +

we have, successively,

n =

(lbm~lbm~.2...blbo)2,

= (0

b,pl

b,p2..

b0)2 ,

21 =

(b,p,

bmp2.. .

b. 0)2,

21+

1 =

(b,p,

bmp2.. .

b. 1

J(n)

(bm-1

brn-2..

.bl

brn)z.

(The last step follows because J(n) =

2l.+

1 and because

= 1.) We have

proved that

J((bmbm--l

...bl

b0)2)

(brn-1

...bl

bobml2;

(1.10)

12 RECURRENT PROBLEMS

that is, in the lingo of computer programming, we get J(n) from n by doing

a one-bit cyclic shift left! Magic. For example, if n = 100 = (1

lOOlOO)

then

J(n)

J((1100100)~)

= (1001001)

which is 64 + 8 + 1 = 73. If we had been

working all along in binary notation, we probably would have spotted this

pattern immediately.

If we start with n and iterate the J function m + 1 times, we’re doing

(“iteration” means

m + 1 one-bit cyclic shifts; so, since n is an

(mfl

)-bit number, we might

applying a function

expect to end up with n again. But this doesn’t quite work. For instance

to itself.)

if n = 13 we have J((1101)~) =

(1011)2,

but then J((1011)~) =

(111)~

and

the process breaks down; the 0 disappears when it becomes the leading bit.

In fact, J(n) must always be < n by definition, since J(n) is the survivor’s

number; hence if J(n) < n we can never get back up to n by continuing to

iterate.

Repeated application of J produces a sequence of decreasing values that

eventually reach a “fixed point,” where J(n) = n. The cyclic shift property

makes it easy to see what that fixed point will be: Iterating the function

enough times will always produce a pattern of all l's whose value is

2”(“)

where y(n) is the number of 1 bits in the binary representation of n. Thus,

since Y( 13) = 3, we have

or more

I’s

j(r(.TTi(l3,...))

23-l

= 7;

similarly

8 or more

~((101101101101011)2)...)) = 2"

1 = 1023.

Curiously enough,

if M is a compact

C” n-manifold

1),

there

exists a differen-

Cable immersion of

Luria

r*mm~

’

-IUS,

but true.

intO

R*”

~Ytnl

Let’s return briefly to our first guess, that J(n) = n/2 when n is even.

but not necessarily

into

~2”

vinl-1,

This is obviously not true in general, but we can now determine exactly when

wonder

Jose-

it is true:

phus was secretly

a topologist?

J(n)

n/2,

21+

1 = (2"+1)/2,

f(2”

-2).

If this number

(2”’

2) is an integer, then n = 2” +

will be a solution,

because

will be less than 2m. It’s not hard to verify that

-2 is a multiple

of 3 when m is odd, but not when m is even. (We will study such things

in Chapter 4.) Therefore there are infinitely many solutions to the equation

1.3 THE JOSEPHUS PROBLEM 13

J(n) = n/2, beginning as follows:

n=2m+l

J(n) =

21f

1 = n/2 n (binary)

3 2

10 5

1010

101010

42 170

85 10101010

Notice the pattern in the rightmost column. These are the binary numbers

for which cyclic-shifting one place left produces the same result as ordinary-

shifting one place right (halving).

Looks like Greek

to me.

OK, we understand the J function pretty well; the next step is to general-

ize it. What would have happened if our problem had produced a recurrence

that was something like

(1.8),

but with different constants? Then we might

not have been lucky enough to guess the solution, because the solution might

have been really weird. Let’s investigate’this by introducing constants a,

and y and trying to find a closed form for the more general recurrence

f(1) =

cc;

f(2n) = 2f(n) + fi,

for

(1.11)

f(2n+1)=2f(n)+y,

for

(Our original recurrence had a = 1,

= -1, and y = 1.) Starting with

f (1) = a and working our way up, we can construct the following general

table for small values of n:

f(n)

2a-f

3201

4af3f3

4a+28+

4a+

fi+2y

7 4a

8a+7p

8a+

6fl

+ y

(1.12)

It seems that a’s coefficient is n’s largest power of 2. Furthermore, between

powers of 2,

0’s

coefficient decreases by 1 down to 0 and y’s increases by 1

up from 0. Therefore if we express f(n) in the form

f(n) = A(n) a + B(n)

+ C(n)y ,

(1.13)

14 RECURRENT PROBLEMS

by separating out its dependence on

/3,

and

it seems that

A(n) =

2m;

B(n) =

2”‘-1-L;

(1.14)

C(n) =

Here, as usual, n =

and

2m,

for n 3

It’s not terribly hard to prove (1.13) and (1.14) by induction, but the

Ho/d onto your

calculations are messy and uninformative. Fortunately there’s a better way

hats, this next part

to proceed, by choosing particular values and then combining them. Let’s

is new stuff.

illustrate this by considering the special case a =

= y = 0, when f(n) is

supposed to be equal to A(n): Recurrence

(1.11)

becomes

A(1) = 1;

A(2n) =

2A(‘n), for

A(2n + 1) = 2A(n),

for

Sure enough, it’s true (by induction on m) that A(2” +

2m.

Next, let’s use recurrence

(1.11)

and solution (1.13) in

Teverse,

by start-

ing with a simple function f(n) and seeing if there are any constants

(OL,

that will define it. Plugging in the constant function f(n) = 1 says that

neat

idea!

1 = a;

1 = 2.1+p;

1 = 2.1+y;

hence the values (a,

= (1, -1, -1) satisfying these equations will yield

A(n)

B(n)

C(n) = f(n) = 1. Similarly, we can plug in f(n) = n:

1 =

2n = 2+n+

L3;

2n+l

2.n+y;

These equations hold for all n when a =

= 0, and y = 1, so we don’t

need to prove by induction that these parameters will yield f(n) = n. We

already know that f(n) = n will be the solution in such a case, because the

recurrence

(1.11)

uniquely defines f(n) for every value of

And now we’re essentially done! We have shown that the functions A(n),

B(n), and C(n) of (1.13), which solve

(1.11)

in general, satisfy the equations

A(n) =

2”)

where n = 2” +

and 0 6

< 2”;

A(n) -B(n)

C(n) = 1

;

A(n) + C(n) = n.

Beware: The au-

thors are expecting

us to figure out

the idea of the

repertoire method

from

seat-of-the-

pants examples,

instead of giving

us a top-down

presentation. The

method works best

with recurrences

that are ‘linear” in

the sense that /heir

solutions can be

expressed as a sum

of arbitrary param-

eters multiplied by

functions of n, as

in (1.13). Equation

(1.13) is the key.

(‘relax = ‘destroy’)

I think I get it:

The binary repre-

sentations of A(n),

B(n), and C(n)

have 1

‘s

in different

positions.

1.3 THE JOSEPHUS PROBLEM 15

Our conjectures in (1.14) follow immediately, since we can solve these equa-

tions to get C(n) = n

A(n) =

and B(n) = A(n)

C(n) = 2”

This approach illustrates a surprisingly useful repertoire method for solv-

ing recurrences. First we find settings of general parameters for which we

know the solution; this gives us a repertoire of special cases that we can solve.

Then we obtain the general case by combining the special cases. We need as

many independent special solutions as there are independent parameters (in

this case three, for

01,

J3,

and y). Exercises 16 and 20 provide further examples

of the repertoire approach.

We know that the original J-recurrence has a magical solution, in binary:

J(bn

bm-1 . . . bl bob) = (bm-1 . . . b,

b,)z ,

where

Does the generalized

Josephus

recurrence admit of such magic?

Sure, why not? We can rewrite the generalized recurrence (1.11) as

f(1) = a;

f(2n

+ j) = 2f(n) +

J3j

for j =

0,l

and n 3

(1.15)

if we let

and

J31

And this recurrence unfolds, binary-wise:

f(bnbm-1

. . .

bob)

2f((bm

b-1

. . .

12) + fib0

4f((b,

b,el

. .

Wz)

2f’b,

+ fib‘,

2mf((bmh)

+2m-1Pbmm,

+.“+@b,

(3bo

= 2”(x +

293b,m,

+ “’ +

2(&q

Suppose we now relax the radix 2 notation to allow arbitrary digits instead

of just 0 and

The derivation above tells us that

f((bm

b-1

. .

bob)

(01

fib,-,

Pb,,mz . . . @b, f’bo

(1.16)

Nice. We would have seen this pattern earlier if we had written (1.12) in

anot her way:

16 RECURRENT PROBLEMS

For example, when n = 100 = (1100100)~, our original

Josephus

values

LX=],

/3=-l,andy=l

yield

n= (1 1 0 0 1 0 O)L

100

f(n) = ( 1 1 -1 -1 1 -1

-1)1

=+64+32-16-8+4-2-l

as before. The cyclic-shift property follows because each block of binary digits

(10 . . .

00)~ in the representation of n is transformed into

(l-l . . .

-l-l)2

= (00 ..,Ol)z.

So our change of notation has given us the compact solution (1.16) to the

There are two

general recurrence (1.15). If we’re really uninhibited we can now generalize

kinds

Ofgenera’-

even more. The recurrence

izations.

One is

cheap and the other

f(i)

for

is valuable.

(1.17)

It is easy to gen-

f(dn + j) = cf(n) +

(3j

forO<j<d

and

n31,

eralize by diluting

a little idea with a

is the same as the previous one except that we start with numbers in radix d

big terminology.

and produce values in radix c. That is, it has the radix-changing solution

It is much more

dificult to pre-

pare a refined and

b-1

. .

.bl

b&i)

cab,

f’b,m,

fib,->

. . . bb,

(3bo)c.

(1.18)

condensed extract

from

several good

For example, suppose that by some stroke of luck we’re given the recurrence

ingredients.

Pdlya

12381

f(1) = 34,

f(2) = 5,

f(3n) = lOf(n) + 76,

for n 3

f(3nfl)

= lOf(n)-2, for n 3

f(3n

+2)

lOf(n)+8,

for n 3 1,

and suppose we want to compute f (19). Here we have d = 3 and c = 10. Now

Perhaps this was a

19 = (201)3, and the radix-changing solution tells us to perform a digit-by-

stroke

bad

luck.

digit replacement from radix 3 to radix 10. So the leading 2 becomes a 5, and

the 0 and 1 become 76 and -2, giving

f(19) =

f((201)3)

= (5 76 -2),. = 1258,

which is our answer.

But in general I’m

Thus

Josephus

and the Jewish-Roman war have led us to some interesting

against recurrences

general recurrences.

war.

1 EXERCISES 17

Exercises

Warmups

Please do all the

All horses are the same color; we can prove this by induction on the

warmups

all the

chapters!

number of horses in a given set. Here’s how: “If there’s just one horse

The

h4gm

‘t

then it’s the same color as itself, so the basis is trivial. For the induction

step, assume that there are n horses numbered 1 to n. By the induc-

tion hypothesis, horses 1 through n

1 are the same color, and similarly

horses 2 through n are the same color. But the middle horses, 2 through

1, can’t change color when they’re in different groups; these are

horses, not chameleons. So horses 1 and n must be the same color as

well, by transitivity. Thus all n horses are the same color; QED.” What,

if anything, is wrong with this reasoning?

2 Find the shortest sequence of moves that transfers a tower of n disks

from the left peg A to the right peg B, if direct moves between A and B

are disallowed. (Each move must be to or from the middle peg. As usual,

a larger disk must never appear above a smaller one.)

Show that, in the process of transferring a tower under the restrictions of

the preceding exercise, we will actually encounter every properly stacked

arrangement of n disks on three pegs.

4 Are there any starting and ending configurations of n disks on three pegs

that are more than 2”

1 moves apart, under Lucas’s original rules?

5 A “Venn diagram” with three overlapping circles is often used to illustrate

the eight possible subsets associated with three given sets:

Can the sixteen possibilities that arise with four given sets be illustrated

by four overlapping circles?

6 Some of the regions defined by n lines in the plane are infinite, while

others are bounded. What’s the maximum possible number of bounded

regions?

7 Let H(n) = J(n+ 1)

J(n). Equation (1.8) tells us that H(2n) = 2, and

H(2n+l)

J(2n+2)-J(2n+l)

(2J(n+l)-l)-(2J(n)+l)

2H(n)-2,

for all n 3 1. Therefore it seems possible to prove that H(n) = 2 for all n,

by induction on n. What’s wrong here?

18 RECURRENT PROBLEMS

Homework exercises

8 Solve the recurrence

Qo = 0~;

Qn =

Qn-l)/Qn-2,

for n > 1.

Assume that Q,, # 0 for all n 3 0. Hint: QJ = (1 +

oc)/(3.

9 Sometimes it’s possible to use induction backwards, proving things from

now that’s a

n to n

1 instead of vice versa! For example, consider the statement

horse of a different

color.

P(n) :

. .

.x,

(

. . +

)

ifxr

,...,

x,30.

This is true when n = 2, since

(x1

+xJ)~

-4~1x2

(x1

-xz)~

3 0.

By setting x,, =

(XI

...

x,~l)/(n

l),

prove that P(n) im-

plies P(n

1) whenever n >

b Show that P(n) and P(2) imply P(2n).

Explain why this implies the truth of P(n) for all

Let Q,, be the minimum number of moves needed to transfer a tower of

n disks from A to B if all moves must be clockwise-that is, from A

to B, or from B to the other peg, or from the other peg to A. Also let R,

be the minimum number of moves needed to go from B back to A under

this restriction. Prove that

Qn=

;;,,,+l

{

;;;,;i

Rn=

+Qnp,+,,

;;;,;’

(You need not solve these recurrences; we’ll see how to do that in Chap-

ter 7.)

11 A Double Tower of Hanoi contains 2n disks of n different sizes, two of

each size. As usual, we’re required to move only one disk at a time,

without putting a larger one over a smaller one.

How many moves does it take to transfer a double tower from one

peg to another, if disks of equal size are indistinguishable from each

other?

b What if we are required to reproduce the original top-to-bottom

order of all the equal-size disks in the final arrangement? [Hint:

This is difficult-it’s really a “bonus problem.“]

12 Let’s generalize exercise lla even further, by assuming that there are

m different sizes of disks and exactly

disks of size k. Determine

Nnl,.

. . ,

n,), the minimum number of moves needed to transfer a tower

when equal-size disks are considered to be indistinguishable.

1 EXERCISES 19

What’s the maximum number of regions definable by n zig-zag lines,

zzz=12

Good luck

keep-

ing the

cheese in

position.

each of which consists of two parallel infinite half-lines joined by a straight

segment?

14 How many pieces of cheese can you obtain from a single thick piece by

making five straight slices? (The cheese must stay in its original position

while you do all the cutting, and each slice must correspond to a plane

in 3D.) Find a recurrence relation for P,, the maximum number of three-

dimensional regions that can be defined by n different planes.

15 Josephus had a friend who was saved by getting into the next-to-last

position. What is I(n), the number of the penultimate survivor when

every second person is executed?

16 Use the repertoire method to solve the general four-parameter recurrence

g(l) =

gVn+j)

h(n)

+w+

Pi,

for j = 0,l and n 3 1.

Hint: Try the function g(n) = n.

Exam problems

If W, is the minimum number of moves needed to transfer a tower of n

disks from one peg to another when there are four pegs instead of three,

show that

Wn(n+1

j/2 6

34’n(n-1

i/2 +

for n > 0.

this like a

five-star general

recurrence?

(Here

T,,

= 2”

1 is the ordinary three-peg number.) Use this to find a

closed form f(n) such that W,(,+r~,~ 6 f(n) for all n 3 0.

18 Show that the following set of n bent lines defines

regions, where

is defined in (1.7): The jth bent line, for 1 < j 6 n, has its zig at (nZi,O)

and goes up through the points (n’j

nj, 1) and (n’j

nn,

1).

19 Is it possible to obtain

regions with n bent lines when the angle at

each zig is 30”?

20 Use the repertoire method to solve the general five-parameter recurrence

h(l) = a;

h(2n + i) = 4h(n) +

yin

(3j

forj=O,l

and

n>l.

Hint: Try the functions h(n) = n and h(n) = n2.

20 RECURRENT PROBLEMS

21 Suppose there are 2n people in a circle; the first n are “good guys”

and the last n are “bad guys!’ Show that there is always an integer m

(depending on n) such that, if we go around the circle executing every

mth person, all the bad guys are first to go. (For example, when n = 3

we can take m = 5; when n = 4 we can take m = 30.)

Bonus problems

22 Show that it’s possible to construct a Venn diagram for all 2” possible

subsets of n given sets, using n convex polygons that are congruent to

each other and rotated about a common center.

23 Suppose that

Josephus

finds himself in a given position j, but he has a

chance to name the elimination parameter q such that every qth person

is executed. Can he always save himself?

Research problems

24 Find all recurrence relations of the form

x _

ao+alX,-1

+...+akXnPk

X,-i + . . +

bkXn-k

whose solution is periodic.

Solve infinitely many cases of the four-peg Tower of Hanoi problem by

proving that equality holds in the relation of exercise 17.

26 Generalizing exercise 23, let’s say that a

Josephus

subset of {1,2,. . . , n}

is a set of k numbers such that, for some

the people with the other n-k

numbers will be eliminated first. (These are the k positions of the “good

guys”

Josephus

wants to save.) It turns out that when n = 9, three of the

possible subsets are non-Josephus, namely

{1,2,5,8,9},

{2,3,4,5,

S},

and

{2,5,6,7,

S}.

There are 13 non-Josephus sets when n = 12, none for

any other values of n 6 12. Are non-Josephus subsets rare for large n?

Yes, and well done

if you find them.

Sums

SUMS ARE EVERYWHERE in mathematics, so we need basic tools to handle

them. This chapter develops the notation and general techniques that make

summation user-friendly.

2.1 NOTATION

In Chapter 1 we encountered the sum of the first n integers, which

wewroteoutas1+2+3+...+(n-1)fn.

The‘...‘insuchformulastells

us to complete the pattern established by the surrounding terms. Of course

we have to watch out for sums like 1

7 + . . .

41.7, which are meaningless

without a mitigating context. On the other hand, the inclusion of terms like

3 and (n

1) was a bit of overkill; the pattern would presumably have been

clear if we had written simply 1 + 2 + . . . + n. Sometimes we might even be

so bold as to write just 1

f..

. + n.

We’ll be working with sums of the general form

a2 +

...

a,,

(2.1)

where each

is a number that has been defined somehow. This notation has

the advantage that we can “see” the whole sum, almost as if it were written

out in full, if we have a good enough imagination.

A term is how long

Each element

of a sum is called a term. The terms are often specified

this course lasts.

implicitly as formulas that follow a readily perceived pattern, and in such cases

we must sometimes write them in an expanded form so that the meaning is

clear. For example, if

+2+

. . .

+2+'

is supposed to denote a sum of n terms, not of 2”-‘, we should write it more

explicitly as

. .

2n-'.

22 SUMS

The three-dots notation has many uses, but it can be ambiguous and a

“Le

signe

,T~~~

bit long-winded. Other alternatives are available, notably the delimited form

indique

/‘on

doit

dormer

k=l

au nombre entier i

(2.2)

to&es ses valeurs

1,2,3

,...,

prendre la somme

which is called Sigma-notation because it uses the Greek letter

(upper-

des termes.”

case sigma). This notation tells us to include in the sum precisely those

J. Fourier

I1021

terms

whose index k is an integer that lies between the lower and upper

limits 1 and n, inclusive. In words, we

“sum

over k, from 1 to n.” Joseph

Fourier introduced this delimited t-notation in 1820, and it soon took the

mathematical world by storm.

Incidentally, the quantity after

(here ok) is called the

summa&.

The index variable k is said to be bound to the

(2.2),

because

the k in ok is unrelated to appearances of k outside the Sigma-notation. Any

other letter could be substituted for k here without changing the meaning of

Well, I wouldn’t

(2.2). The letter i is often used (perhaps because it stands for “index”), but

want

use

we’ll generally sum on k since it’s wise to keep i for

&i.

as the index vari-

able instead of k in

It turns out that a generalized Sigma-notation is even more useful than

(2.2);

those

letters

the delimited form: We simply write one or more conditions under the

x.,

are “free variables”

to specify the set of indices over which summation should take place. For

that

have

mean-

example, the sums in (2.1) and (2.2) can also be written as

outside the

here.

ak .

(2.3)

l<k<n

In this particular example there isn’t much difference between the new form

and

(2.2),

but the general form allows us to take sums over index sets that

aren’t restricted to consecutive integers. Fbr example, we can express the sum

of the squares of all odd positive integers below 100 as follows:

l<k<lOO

k odd

The delimited equivalent of this sum,

2k + 1)’ ,

k=O

is more cumbersome and less clear. Similarly, the sum of reciprocals of all

prime numbers between 1 and N is

;;

P<N

p prime

2.1 NOTATION 23

the delimited form would require us to write

The summation

where

denotes the kth prime and n(N) is the number of primes < N.

(Incidentally, this sum gives the approximate average number of distinct prime

factors of a random integer near N, since about 1 /p of those integers are

divisible by p. Its value for large N is approximately lnln N + 0.261972128;

stands for the natural logarithm of x, and In In x stands for ln( In x)

The biggest advantage of general Sigma-notation is that we can manip-

ulate it more easily than the delimited form. For example, suppose we want

symbol looks like

a distorted pacman.

to change the index variable k to k + 1. With the general form, we have

ak+l

;

l<k<n l<k+l<n

it’s easy to see what’s going on, and we can do the substitution almost without

thinking. But with the delimited form, we have

n--l

tak+1;

k=l

k=O

A tidy sum.

it’s harder to see what’s happened, and we’re more likely to make a mistake.

On the other hand, the delimited form isn’t completely useless. It’s

nice and tidy, and we can write it quickly because (2.2) has seven symbols

compared with (2.3)‘s eight. Therefore we’ll often use

with upper and

lower delimiters when we state a problem or present a result, but we’ll prefer

to work with relations-under-x when we’re manipulating a sum whose index

variables need to be transformed.

That’s nothing.

The

sign occurs more than 1000 times in this book, so we should be

You should see how

many times

ap-

sure that we know exactly what it means. Formally, we write

pears in The Iliad.

(2.4)

Pikl

as an abbreviation for the sum of all terms

such that k is an integer

satisfying a given property P(k). (A “property P(k)” is any statement about

k that can be either true or false.) For the time being, we’ll assume that

only finitely many integers k satisfying P(k) have

# 0; otherwise infinitely

many nonzero numbers are being added together, and things can get a bit

tricky. At the other extreme, if P(k) is false for all integers k, we have an

“empty” sum; the value of an empty sum is defined to be zero.

24 SUMS

A slightly modified form of (2.4) is used when a sum appears within the

text of a paragraph rather than in a displayed equation: We write

‘x.pCkl

ak’,

attaching property P(k) as a subscript of

so that the formula won’t stick

out too much. Similarly, ‘xF=,

ak’

is a convenient alternative to (2.2) when

we want to confine the notation to a single line.

People are often tempted to write

n-1

k(k-

l)(n-

k) instead of

k(k- l)(n- k)

k=2 k=O

because the terms for k = 0, 1, and n in this sum are zero. Somehow it

seems more efficient to add up n

2 terms instead of n + 1 terms. But such

temptations should be resisted; efficiency of computation is not the same as

efficiency of understanding! We will find it advantageous to keep upper and

lower bounds on an index of summation as simple as possible, because sums

can be manipulated much more easily when the bounds are simple. Indeed,

the form

EL!;

can even be dangerously ambiguous, because its meaning is

not at all clear when n = 0 or n = 1 (see exercise 1). Zero-valued terms cause

no harm, and they often save a lot of trouble.

So far the notations we’ve been discussing are quite standard, but now

we are about to make a radical departure from tradition. Kenneth Iverson

introduced a wonderful idea in his programming language APL

[161,

page

111,

and we’ll see that it greatly simplifies many of the things we want to do in

this book. The idea is simply to enclose a true-or-false statement in brackets,

and to sav that the result is 1 if the statement is true. 0 if the statement is

Hev: The

“Kro-

false. For example,

neiker delta”

that

I’ve seen in other

if p is a prime number;

books

mean

prime] =

if p is not a prime number.

6k,,

, which is 1 if

k=n,

Ooth-

erwise) is just a

Iverson’s convention allows us to express sums with no constraints whatever

special case

on the index of summation, because we can rewrite (2.4) in the form

lverson

‘s

conven-

tion: We can write

[P(k)]

(2.5)

[ k = n

]

instead.

If P(k) is false, the term ok[P(k)]

is zero, so we can safely include it among

the terms being summed. This makes it easy to manipulate the index of

summation, because we don’t have to fuss with boundary conditions.

A slight technicality needs to be mentioned: Sometimes ok isn’t defined

for all integers k. We get around this difficulty by assuming that [P(k)] is

“very strongly zero” when P(k) is false; it’s so much zero, it makes ok [P(k)]

equal to zero even when ok is undefined. For example, if we use Iverson’s

2.1 NOTATION 25

convention to write the sum of reciprocal primes $ N as

prime1

there’s no problem of division by zero when p = 0, because our convention

tells us that

prime]

< Nl/O = 0.

Let’s sum up what we’ve discussed so far about sums. There are two

good ways to express a sum of terms: One way uses ‘. . .‘, the other uses

‘

‘.

The three-dots form often suggests useful manipulations, particularly

the combination of adjacent terms, since we might be able to spot a simplifying

pattern if we let the whole sum hang out before our eyes. But too much detail

can also be overwhelming. Sigma-notation is compact, impressive to family

. .

and it’s less

and friends, and often suggestive of manipulations that are not obvious in

likely to lose points

on an exam for

three-dots form. When we work with Sigma-notation, zero terms are not

“lack of rigor.”

generally harmful; in fact, zeros often make t-manipulation easier.

2.2 SUMS AND RECURRENCES

OK, we understand now how to express sums with fancy notation.

But how does a person actually go about finding the value of a sum? One way

is to observe that there’s an intimate relation between sums and recurrences.

The sum

(Think of S, as is equivalent to the recurrence

not just a single

number, but as a

sequence defined for

= ao;

all n 3 0

S-1

+ a,,

for n > 0.

(2.6)

Therefore we can evaluate sums in closed form by using the methods we

learned in Chapter 1 to solve recurrences in closed form.

For example, if a,, is equal to a constant plus a multiple of n, the sum-

recurrence (2.6) takes the following general form:

Ro=cx;

R,=R,-l+B+yn,

for n > 0.

Proceeding as in Chapter 1, we find

= a +

+ 26 + 37, and

so on; in general the solution can be written in the form

R, = A(n)

+ B(n)

C(n)y

(2.8)

26 SUMS

where A(n), B(n), and C(n) are the coefficients of dependence on the general

parameters

01,

and

The repertoire method tells us to try plugging in simple functions of n

for

R,,

hoping to find constant parameters

01,

(3,

and y where the solution is

especially simple. Setting R, = 1 implies

= 1,

= 0, y = 0; hence

A(n) = 1.

Setting R, = n implies a = 0,

y = 0; hence

B(n) = n.

Setting R, =

implies a = 0,

= -1, y = 2; hence

2C(n) -B(n) =

and we have C(n) =

(n2

+n)/2.

Easy as pie.

Therefore if we wish to evaluate

a + bk) ,

k=O

the sum-recurrence (2.6) boils down to (2.7) with a =

= a, y = b, and the

answer is

+ aB(n) + bC(n) = a(n + 1) + b(n +

l)n/2.

Conversely, many recurrences can be reduced to sums; therefore the spe-

cial methods for evaluating sums that we’ll be learning later in this chapter

will help us solve recurrences that might otherwise be difficult. The Tower of

Hanoi recurrence is a case in point:

To = 0;

T,,

= 2T,_,

for n > 0.

It can be put into the special form (2.6) if we divide both sides by 2”:

To/2' = 0;

TJ2"

= T,-,/2-' +l/2n, for n > 0.

Now we can set

= T,/2n, and we have

so = 0;

= s,~-’

+2-n)

for n > 0.

Actually easier;

14n+1)14n+3) .

It follows that

t2-k

k=l

2.2 SUMS AND RECURRENCES 27

(Notice that we’ve left the term for k = 0 out of this sum.) The sum of the

geometricseries2~‘+2~2+~~~+2~“=(~)’+(~)2+~~~+(~)nwillbederived

later in this chapter; it turns out to be 1

)“. Hence

T,,

= 2”S, = 2”

We have converted T, to

in this derivation by noticing that the re-

currence could be divided by

2n.

This trick is a special case of a general

technique that can reduce virtually any recurrence of the form

a,T,, =

bnTn-1

(2.9)

to a sum. The idea is to multiply both sides by a summation factor, s,:

s,a,T,,

s,,bnTn-1 + snc,, .

This factor s, is cleverly chosen to make

n n

h-1

an-l

Then if we write

s,a,T,,

we have a sum-recurrence,

Sn-1

+SnCn.

Hence

= socuT +

skck

s.lblTo

skck

k=l k=l

and the solution to the original recurrence (2.9) is

ha,

s,b,To +

&Ck

k=l

(2.10)

[The

value

cancels out, so it

can

anything

but zero.)

For example, when n = 1 we get

(s~b,To

+slcl)/slal

(b,To

+cl)/al.

But how can we be clever enough to find the right s,? No problem: The

relation

s,,

snPl

anPI

/b,

can be unfolded to tell us that the fraction

a,-

a,-2..

. al

b,bnp,...bz

’

(2.11)

or any convenient constant multiple of this value, will be a suitable summation

factor. For example, the Tower of Hanoi recurrence has a,, = 1 and b, = 2;

the general method we’ve just derived says that

2-”

is a good thing to

multiply by, if we want to reduce the recurrence to a sum. We don’t need a

brilliant flash of inspiration to discover this multiplier.

We must be careful, as always, not to divide by zero. The summation-

factor method works whenever all the a’s and all the b’s are nonzero.

28 SUMS

Let’s apply these ideas to a recurrence that arises in the study of “quick-

sort,” one of the most important methods for sorting data inside a computer.

(Quicksort was

The average number of comparison steps made by quicksort when it is applied

invented

H0arc

to n items in random order satisfies the recurrence

in 1962 [158].)

k=O

for n > 0.

(2.12)

Hmmm. This looks much scarier than the recurrences we’ve seen before; it

includes a sum over all previous values, and a division by n. Trying small

cases gives us some data (Cl = 2, Cl = 5,

= T) but doesn’t do anything

to quell our fears.

We can, however, reduce the complexity of

(2.12)

systematically, by first

getting rid of the division and then getting rid of the

sign. The idea is to

multiply both sides by n, obtaining the relation

n-1

nC,

n2+n+2xCk,

for n > 0;

k=O

hence, if we replace n by n

n-2

(n-l)cnpj

(n-1)2+(n-1)+2xck,

forn-1

>O.

k=O

We can now subtract the second equation from the first, and the

sign

disappears:

nC,

1)&l

= 2n + 2C,-1 ,

for n >

It turns out that this relation also holds when n =

because Cl = 2. There-

fore the original recurrence for

reduces to a much simpler one:

co = 0;

nC,

(n + 1 )C,-I + 2n,

for n > 0.

Progress. We’re now in a position to apply a summation factor, since this

recurrence has the form of (2.9) with a, = n,

= n + 1, and c, = 2n.

The general method described on the preceding page tells us to multiply the

recurrence through by some multiple of

a,._1

an-l.

. .

(n-l).(n-2).....1

b,b,-,

. . b2 =

(n+l).n...:3

(n+l)n

2.2 SUMS AND RECURRENCES 29

We started with a

in the recur-

rence, and worked

hard to get rid of

it.

But then after ap-

plying

a summation

factor, we

came up

with another

Are sums good, or

bad, or what?

But your spelling is

a/wrong.

The solution, according to

(2.10),

is therefore

= 2(n + 1)

k=l

k+l

The sum that remains is very similar to a quantity that arises frequently

in applications. It arises so often, in fact, that we give it a special name and

a special notation:

,+;+...+;

f;.

k=l

(2.13)

The letter H stands for “harmonic”;

is a harmonic number, so called

because the kth harmonic produced by a violin string is the fundamental

tone produced by a string that is l/k times as long.

We can complete our study of the quicksort recurrence

(2.12)

by putting

into closed form; this will be possible if we can express

in terms of H,.

The sum in our formula for

We can relate this to

without much difficulty by changing k to k

1 and

revising the boundary conditions:

(

>--

1 1

1+nSi=

H,-5.

l<k<n

nfl

Alright! We have found the sum needed to complete the solution to

(2.12):

The average number of comparisons made by quicksort when it is applied to

n randomly ordered items of data is

2(n+l)H,-2n.

(2.14)

As usual, we check that small cases are correct:

= 0, Cl = 2,

= 5.

30 SUMS

2.3 MANIPULATION OF SUMS

Not to be confused

with finance.

The key to success with sums is an ability to change one

into

another that is simpler or closer to some goal. And it’s easy to do this by

learning a few basic rules of transformation and by practicing their use.

Let K be any finite set of integers. Sums over the elements of K can be

transformed by using three simple rules:

pk;

cak

(distributive law)

(2.15)

kEK

~iak+bk)

&+~bk;

(associative law) (2.16)

kEK

%(k)

(commutative law)

(2.17)

kEK

p(k)EK

The distributive law allows us to move constants in and out of a

The

associative law allows us to break a

into two parts, or to combine two

x’s

into one. The commutative law says that we can reorder the terms in any way

we please; here p(k) is any permutation of the set of all integers. For example,

Why not call it

if K = (-1 (0, +l} and if p(k) =

-k,

these three laws tell us respectively that

permutative instead

of commutative?

ca-1 +

cao

+ cal =

c(a-j

faofal);

(distributive law)

(a-1

Sb-1)

(ao+b)

+ (al

+bl)

(a-l+ao+al)+(b-l+bo+bl);

(associative law)

a-1 + a0 + al = al + a0 + a-1 .

(commutative law)

Gauss’s trick in Chapter 1 can be viewed

an application of these three

basic laws. Suppose we want to compute the general sum of an arithmetic

progression,

S =

(afbk).

O<k$n

By the commutative law we can replace k by n

k, obtaining

S =

(a+b(n-k)) =

(a+bn-bk).

O<n-k<n

O<k<n

These two equations can be added by using the associative law:

This is something

like changing vari-

ables inside an

integral, but easier.

2S =

((a+bk)+(a+bn-bk)) =

(2afbn).

O<k<n

O<k$n

“What’s one

and one and one

and one?”

“1

don’t know,”

said Alice.

‘7

lost count.”

“She can’t do

Addition.”

-Lewis Carroll

[44]

Additional, eh?

2.3 MANIPULATION OF SUMS 31

And we can now apply the distributive law and evaluate a trivial sum:

2S =

(2a+bn)

1 =

(2a+bn)(n+l).

O<k<n

Dividing by 2, we have proved that

a+bk) =

(a+ibn)(n+l).

(2.18)

k=O

The right-hand side can be remembered as the average of the first and last

terms, namely

(a + (a + bn)), times the number of terms, namely (n + 1).

It’s important to bear in mind that the function p(k) in the general

commutative law (2.17) is supposed to be a permutation of all the integers. In

other words, for every integer n there should be exactly one integer k such that

p(k) = n. Otherwise the commutative law might fail; exercise 3 illustrates

this with a vengeance. Transformations like p(k) = k + c or p(k) = c

where c is an integer constant, are always permutations, so they always work.

On the other hand, we can relax the permutation restriction a little bit:

We need to require only that there be exactly one integer k with p(k) = n

when n is an element of the index set K. If n

K (that is, if n is not in K),

it doesn’t matter how often p(k) = n occurs, because such k don’t take part

in the sum. Thus, for example, we can argue that

a2k

a2k,

(2.19)

kEK

WSK

2kEK 2kEK

k even

n even

2k even

since there’s exactly one k such that 2k = n when n

K and n is even.

Iverson’s convention, which allows us to obtain the values 0 or 1 from

logical statements in the middle of a formula, can be used together with the

distributive, associative, and commutative laws to deduce additional proper-

ties of sums. For example, here is an important rule for combining different

sets of indices: If K and K’ are any sets of integers, then

ak.

kE:K

kEK’ kEKnK’ kEKuK’

This follows from the general formulas

(2.20)

ak[kEK]

(2.21)

kEK

and

[kEK]+[kEK’]

[kEKnK’]+[kEKuK’].

(2.22)

32 SUMS

Typically we use rule (2.20) either to combine two almost-disjoint index sets,

as in

n n

tak

ak,

for 1 < m < n;

k=l

k=m

k=l

or to split off a single term from a sum, as in

ak ,

for n 3 0.

O<k<n

I<k<n

This operation of splitting off a term is the basis of a perturbation

method that often allows us to evaluate a sum in closed form. The idea

is to start with an unknown sum and call it S,:

ak.

O<k<n

(Name and conquer.) Then we rewrite

Sn+l

in two ways, by splitting off both

its last term and its first term:

S,+

an+1

O<k<n+l

ik$n+l

a0+

ak+l

l<k+lSn+l

ak+l

(2.24)

O<k<n

Now we can work on this last sum and try to express it in terms of S,. If we

succeed, we obtain an equation whose solution is the sum we seek.

For example, let’s use this approach to find the sum of a general geomet-

If it’s geometric,

ric progression,

there should be a

geometric proof.

S, =

axk.

04kSn

The general perturbation scheme in (2.24) tells us that

S, +

axn+’

ax0

axk+’

O<k<n

and the sum on the right is xxobkGn

axk

= xS, by the distributive law.

Therefore S, +

ax”+’

= a +

xSnr

and we can solve for S, to obtain

Laxk

aycJxi+‘,

forx#l

k=O

(2.25 )

2.3 MANIPULATION OF SUMS 33

(When x = 1, the sum is of course simply (n + 1 )a.) The right-hand side

Ah yes, this formula

can be remembered as the first term included in the sum minus the first term

was drilled into me

in high school.

excluded (the term after the last), divided by 1 minus the term ratio.

That was almost too easy. Let’s try the perturbation technique on a

slightly more difficult sum,

k2k

O<k<n

In this case we have

= 0,

= 2, Sl = 10,

= 34,

= 98; what is the

general formula? According to (2.24) we have

S,+(n+1)2”+’

(k+1)2k+‘;

O<k<n

so we want to express the right-hand sum in terms of S,. Well, we can break

it into two sums with the help of the associative law,

k2k+’

2k+‘,

O$k<n O<k<n

and the first of the remaining sums is

2S,.

The other sum is a geometric

progression, which equals (2

2”+2)/( 1

2) =

2n+2

2 by (2.25). Therefore

we have

+ (n + 1

)2n+’

= 2S, +

2n+2

2, and algebra yields

k2k = (n-

1)2"+'

+2.

O<k<n

Now we understand why

= 34: It’s 32 + 2, not 2.17.

A similar derivation with x in place of 2 would have given us the equation

S,+(n+

1)x"+'

=x&+(x-xXn+'

)/(l

x); hence we can deduce that

kxk =

x-(nt

l)xn+'

+nxn+2

-x)2

for x # 1

k=O

(2.26)

It’s interesting to note that we could have derived this closed form in a

completely different way, by using elementary techniques of differential cal-

culus. If we start with the equation

Xn+l

l-x

k=O

and take the derivative of both sides with respect to x, we get

k&’

(1-x)(-(n+l)xn)+l-xn+'

1 -(n+

l)xn

+nxn+’

k=O

-x)2

34 SUMS

because the derivative of a sum is the sum of the derivatives of its terms. We

will see many more connections between calculus and discrete mathematics

in later chapters.

2.4 MULTIPLE SUMS

The terms of a sum might be specified by two or more indices, not

just by one. For example, here’s a double sum of nine terms, governed by two

Oh no, a nine-term

indices

and k:

governor.

Cljbk

= olbl + olb2 + olb3

l<j,k<3

+ azbl +

a2b2

azb3

+ a3bl +

a3b2

+ a3b3.

We use the same notations and methods for such sums as we do for sums with

a single index. Thus, if P(j, k) is a property of j and k, the sum of all terms

oj,k

such that P(j, k) is true can be written in two ways, one of which uses

Iverson’s convention and sums over all pairs of integers j and k:

aj,k =

aj,k [P(i,k)] .

PLj,kl

i,k

Notice that this

doesn’t mean to

sum over

all

j 3 1

and all k < 3.

Only one

sign is needed, although there is more than one index of sum-

mation;

denotes a sum over all combinations of indices that apply.

We also have occasion to use two

x’s,

when we’re talking about a sum

of sums. For example,

aj,k

[P(j,k)]

is an abbreviation for

t(Faj,k

[Plj.ki])

which is the sum, over all integers j, of

oj,k

[P(j, k)], the latter being the

Multiple C’s are

sum over all integers k of all terms

oj,k

for which P(j, k) is true. In such cases

evaluated right to

we say that the double sum is “summed tist on k!’ A sum that depends on

left (inside-out).

more than one index can be summed first on any one of its indices.

In this regard we have a basic law called interchanging the order of

summation, which generalizes the associative law (2.16) we saw earlier:

aj,k[P(j,k)]

aj,k =

aj,k[P(j,k)].

P(j,k)

(2.27)

2.4 MULTIPLE SUMS 35

The middle term of this law is a sum over two indices. On the left,

stands for summing first on k, then on j. On the right,

stands for

summing first on j, then on k. In practice when we want to evaluate a double

sum in closed form, it’s usually easier to sum it first on one index rather than

on the other; we get to choose whichever is more convenient.

Who’s panicking?

Sums of sums are no reason to panic, but they can appear confusing to

think this rule

is fairly obvious

a beginner, so let’s do some more examples. The nine-term sum we began

compared to some

with provides a good illustration of the manipulation of double sums, because

of the stuff in

that sum can actually be simplified, and the simplification process is typical

Chapter

of what we can do with

x’s:

Cljbk

xCljbk[l

<j,k63]

tCljbk[l

<j<3][1

<k<3]

l<j,k<3

j,k

Cljbk[l

<j<3][1

Sk631

xaj[l

<j<3]tbk[l

<k631

xaj[l

<i631

xbk[l

<k63]

The first line here denotes a sum of nine terms in no particular order. The

second line groups them in threes, (al

+ al

b3)

+ (albl +

a2b2

azb3) + (a3bl +

a3b2

+ a3b3). The third line uses the distributive law to

factor out the a’s, since

and [l 6 j 6

do not depend on k; this gives

al(bl +

b3)

+ az(br +

b3)

+ a3(bl +

+ b3). The fourth line is

the same as the third, but with a redundant pair of parentheses thrown in

SO that the fifth line won’t look so mysterious. The fifth line factors out the

(br

b3)

that occurs for each value of j: (al +

+ as)(b, +

+ b3).

The last line is just another way to write the previous line. This method of

derivation can be used to prove a general distributive law,

valid for all sets of indices J and K.

The basic law (2.27) for interchanging the order of summation has many

variations, which arise when we want to restrict the ranges of the indices

36 SUMS

instead of summing over all integers j and k. These variations come in two

flavors, vanilla and rocky road. First, the vanilla version:

(2.29)

This is just another way to write

(2.27),

since the Iversonian

J, kE K]

factors into

K]. The vanilla-flavored law applies whenever the ranges

of j and k are independent of each other.

The rocky-road formula for interchange is a little trickier. It applies when

the range of an inner sum depends on the index variable of the outer sum:

ai,k

ai,k.

jEJ kEK(j)

M’K’ iEJ’(k)

Here the sets J, K(j), K’, and J’(k) must be related in such a way that

(2.30)

[jEJl[kEK(j)] = [kEK’l[jEJ’(k)].

A factorization like this is always possible in principle, because we can let

J = K’ be the set of all integers and K(j) = J’(k) be the basic property P(j, k)

that governs a double sum. But there are important special cases where the

sets J, K(j), K’, and J’(k) have a simple form. These arise frequently in

applications. For example, here’s a particularly useful factorization:

[16j<nl[j<k<nl

[l<j<k<nl

[l<k<nl[l<j<kI.

(2.31)

This Iversonian equation allows us to write

aj,k

aj,k.

j=l

k=j

l<j<k<n

k=l

j=l

One of these two sums of sums is usually easier to evaluate than the other;

(Now is a

good

we can use (2.32) to switch from the hard one to the easy one.

time

warmup

Let’s apply these ideas to a useful example. Consider the array

exercises 4 and 6.)

(Or to check out

the Snickers bar

a2al

a2 a2

languishing in the

a3al

a3 a2

freezer.)

products ojok. Our goal will be to find a simple formula

for

2.4 MULTIPLE SUMS 37

Does rocky road

have fudge in it?

the sum of all elements on or above the main diagonal of this array. Because

ojok = okoj, the array is symmetrical about its main diagonal; therefore

will be approximately half the sum of

all

the elements (except for a fudge

factor that takes account of the main diagonal).

Such considerations motivate the following manipulations. We have

Sq =

CljClk =

ClkClj

ajak

Sn,

l<j<k<n

l$k<j<n l<k<j<n

because we can rename (j, k) as (k, j). Furthermore, since

[16j<k<nl+[l<k<j<n]

[l<j,k<n]+[l<j=k<n],

we have

The first sum is (xy=, oj) (xE=, ok) =

ok)‘, by the general distribu-

tive law (2.28). The second sum is

Et=,

at. Therefore we have

(2.33)

an expression for the upper triangular sum in terms of simpler single sums.

Encouraged by such success, let’s look at another double sum:

S =

(ok-Clj)(bk-bj).

l<j<k<n

Again we have symmetry when j and k are interchanged:

(oj-ok)(bj-bk)

(ok-oj)(bk-bj).

l<k<j<n l<k<j$n

So we can add S to itself, making use of the identity

[l<j<k<n]+[l<k<j<n]

[l<j,k<nl-[l<j=kCnl

to conclude that

2s =

(aj

ak)(bj

bk)

(aj

ak)(bj

-bk)

l$j,k<n

$j=k$n

38 SUMS

The second sum here is zero; what about the first? It expands into four

separate sums, each of which is vanilla flavored:

ojbj

ojbk

akbj +

akbk

l~j,k<n

l$j,k~n

l<j,k<n l<j,k<n

okbk

ojbk

l<j,k$n

l<j,k<n

2T-L

Clkbk

l<k<n

In the last step both sums have been simplified according to the general

distributive law (2.28). If the manipulation of the first sum seems mysterious,

here it is again in slow motion:

akbk

akh

l<j,k<n

l$k$n

l<j<n

= 2

okbk

$k<n l<j<n

= 2

okbkn = 2n

okbk.

l<k<:n l<k<n

An index variable that doesn’t appear in the summand (here j) can simply

be eliminated if we multiply what’s left by the size of that variable’s index

set (here n).

Returning to where we left off, we can now divide everything by 2 and

rearrange things to obtain an interesting formula:

(&)@k)

n~akbk-,<&jor-ai)(bibrl.

c2.34)

. ,

This identity yields Chebyshev’s summation inequalities as a special case:

(gok)(gbk)

n&lkbk.

ifo,

<...<o,andbl

6”‘Gbn;

(zok)($bk)

ngakbr,

ifal

6...<oa,andbl

3...abn.

(Chebyshev actu-

ally proved the

analogous result

for integrals

instead of sums:

!.I-:

f(x)

dx)

(J-1:

g(x)

dx)

. (.I-:f(xMx)

dx),

if f(x) and g(x)

are monotone

nondecreasing

functions.)

(In general, if al <

...

< a, and if p is a permutation of (1,. . . , n}, it’s

possible to prove that the largest value of

I;=,

akbPCk)

occurs when b,(l) 6

. . .

bp(n),

and the smallest value occurs when b,(l) 3 . . . 3

b,(,)

2.4 MULTIPLE SUMS 39

Multiple summation has an interesting connection with the general op-

eration of changing the index of summation in single sums. We know by the

commutative law that

a,(k)

p(k)EK

if p(k) is any permutation of the integers. But what happens when we replace

k by f(j), where f is an arbitrary function

f: J

--+

that takes an integer j

J into an integer f(j)

K? The general formula for

index replacement is

Of(j)

ak#f-(k))

(2.35)

jCJ

kEK

where #f-(k) stands for the number of elements in the set

f-(k) = {j If(j) = k> y

that is, the number of values of j

J such that f(j) equals k.

It’s easy to prove (2.35) by interchanging the order of summation,

(h(j)

[f(j)=k]

akt[f(j)=k]

jEJ

jEJ kEK jCJ

since

xjEJ[f(j)

=k] = #f-(k). In the special case that f is a one-to-one

My other math

correspondence between J and K, we have

#f-(k)

= 1 for all k, and the

teacher calls this a

“bijection”; maybe

general formula (2.35) reduces to

171

learn to love

that word some day.

af(j)

xak.

And then again. . .

jEJ f(jlEK

kEK

This is the commutative law (2.17) we had before, slightly disguised.

Our examples of multiple sums so far have all involved general terms like

or bk. But this book is supposed to be concrete, so let’s take a look at a

multiple sum that involves actual numbers:

40 SUMS

The normal way to evaluate a double sum is to sum first on j or first

on k, so let’s explore both options.

s,=

likGn

l$j<k

k-j

summing first on j

replacing j by k

l<k<n

O<j<kbl

simplifying the bounds on j

Hk-1

<k<n

l<k+l$n

by (2.13), the definition of

HkP1

replacing k by k + 1

. simplifying the bounds on k

O<k<n

Alas! We don’t know how to get a sum of harmonic numbers into closed form.

If we try summing first the other way, we get

Get out the whip.

summing first on k

l<j<n

j<k+jin

l<j<n

O<k<n-j

Hn-i

lgjsn

<n-j<n

O$j<n

replacing k by k + j

simplifying the bounds on k

by (2.13), the definition of Hn-j

replacing j by n

simplifying the bounds on j

We’re back at the same impasse.

But there’s another way to proceed, if we replace k by k + j before

deciding to reduce

to a sum of sums:

s,=

l<j<k<n

“j

recopying the given sum

replacing k by k + j

2.4 MULTIPLE SUMS 41

summing first on j

the sum on j is trivial

by the associative law

l<kbn l<k<n

=n by gosh

nH,-n.

by (2.13), the definition of H,

was smart to say

Aha!

We’ve found S,. Combining this with the false starts we made gives us

k 6 n instead of

k < n

1 in this

a further identity as a bonus:

derivation. Simple

bounds save energy.

Hk =

nH,-n

Obk<n

(2.36)

We can understand the trick that worked here in two ways, one algebraic

and one geometric. (1) Algebraically, if we have a double sum whose terms in-

volve k+f( j), where f is an arbitrary function, this example indicates that it’s

a good idea to try replacing k by k-f(j) and summing on j. (2) Geometrically,

we can look at this particular sum

as follows, in the case n = 4:

k=l

k=2

k=3

k=4

j=l

;

j=2

;

j=3

j=4

Our first attempts, summing first on j (by columns) or on k (by rows), gave

+ Hz + HI. The winning idea was essentially to sum

by diagonals, getting

+ 5.

2.5 GENERAL METHODS

Now let’s consolidate what we’ve learned, by looking at a single

example from several different angles. On the next few pages we’re going to

try to find a closed form for the sum of the first n squares, which we’ll call

0,:

k2, for n > 0.

(2.37)

O<k<n

We’ll see that there are at least seven different ways to solve this problem,

and in the process we’ll learn useful strategies for attacking sums in general.

42 SUMS

First, as usual, we look at some small cases.

0123456

0 1 4 9 16 25 36

100

121

144

q l

0 1 5 14 30 55 91

140 204 285 385 506 650

No closed form for

is immediately evident; but when we do find one, we

can use these values as a check.

Method 0: You could look it up.

A problem like the sum of the first n squares has probably been solved

before, so we can most likely find the solution in a handy reference book.

Sure enough, page 72 of the CRC Standard Mathematical Tables

[24]

has the

answer:

n(n+1)(2n+l)

for n 3 0.

(2.38)

Just to make sure we haven’t misread it, we check that this formula correctly

gives

5.6.1

l/6

= 55. Incidentally, page 72 of the CRC Tables has

further information about the sums of cubes, . . . , tenth powers.

The definitive reference for mathematical formulas is the Handbook of

Mathematical Functions, edited by Abramowitz and Stegun

[2].

Pages 813-

(Harder sums

814 of that book list the values of

Cl,,

for n 6 100; and pages 804 and 809

can

found

exhibit formulas equivalent to (2.38), together with the analogous formulas

in Hansen’s

for sums of cubes, . . . , fifteenth powers, with or without alternating signs.

comprehensive

table (1471.)

But the best source for answers to questions about sequences is an amaz-

ing little book called the Handbook of Integer Sequences, by Sloane [270],

which lists thousands of sequences by their numerical values. If you come

up with a recurrence that you suspect has already been studied, all you have

to do is compute enough terms to distinguish your recurrence from other fa-

mous ones; then chances are you’ll find a pointer to the relevant literature in

Sloane’s Handbook. For example,

5, 14, 30, . . . turns out to be Sloane’s

sequence number 1574, and it’s called the sequence of “square pyramidal

numbers” (because there are

El,

balls in a pyramid that has a square base of

balls). Sloane gives three references, one of which is to the handbook of

Abramowitz and Stegun that we’ve already mentioned.

Still another way to probe the world’s store of accumulated mathematical

wisdom is to use a computer program (such as MACSYMA) that provides

tools for symbolic manipulation. Such programs are indispensable, especially

for people who need to deal with large formulas.

It’s good to be familiar with standard sources of information, because

they can be extremely helpful. But Method 0 isn’t really consistent with the

spirit of this book, because we want to know how to figure out the answers

Or, at least to

problems having

the same answers

‘\

2.5 GENERAL METHODS 43

,’

by ourselves. 6he look-up method is limited to problems that other people

have decided are worth considering; a new problem won’t be there.

as problems that

other people have

Method 1: Guess the answer, prove it by induction.

decided to consider.

Perhaps a little bird has told us the answer to a problem, or we have

arrived at a closed form by some other less-than-rigorous means. Then we

merely have to prove that it is correct.

We might, for example, have noticed that the values of

have rather

small prime factors, so we may have come up with formula (2.38) as something

that works for all small values of n. We might also have conjectured the

equivalent formula

n(n+ t)(n+ 1)

’

for n > 0,

(2.39)

which is nicer because it’s easier to remember. The preponderance of the

evidence supports (2.3g), but we must prove our conjectures beyond all rea-

sonable doubt. Mathematical induction was invented for this purpose.

“Well, Your Honor, we know that 00 = 0 =

0(0+~)(0+1)/3,

so the basis

is easy. For the induction, suppose that n > 0, and assume that (2.39) holds

when n is replaced by n

1. Since

we have

3U, = (n- l)(n- t)(n) +

3n2

(n3

in2 + $n) +

3n2

(n3

+ in2 +

in)

= n(n+

t)(n+

1).

Therefore (2.39) indeed holds, beyond a reasonable doubt, for all n >

0.”

Judge Wapner, in his infinite wisdom, agrees.

Induction has its place, and it is somewhat more defensible than trying

to look up the answer. But it’s still not really what we’re seeking. All of

the other sums we have evaluated so far in this chapter have been conquered

without induction; we should likewise be able to determine a sum like

from scratch. Flashes of inspiration should not be necessary. We should be

able to do sums even on our less creative days.

Method 2: Perturb the sum.

So let’s go back to the perturbation method that worked so well for the

geometric progression (2.25). We extract the first and last terms of

I,,+~

44 SUMS

order to get an equation for

0,:

q ,+(n+l)’ =

(k+l)’

(k2+2k+l)

O<k<n

O<k$n

k2+2

O<k<n O<k<n O$k<n

k +

(n+l).

O<k<n

Oops- the

On’s

cancel each other. Occasionally, despite our best efforts, the

perturbation method produces something like

= I&, so we lose.

Seems more

like a

On the other hand, this derivation is not a total loss; it does reveal a way

draw.

to sum the first n integers in closed form,

k =

(n+l)2-(n+l),

O<k<n

even though we’d hoped to discover the sum of first integers squared. Could

it be that if we start with the sum of the integers cubed, which we might

call

we will get an expression for the integers squared? Let’s try it.

GD,+(n+1)3 =

(k+l)3

(k3+3k2+3k+l)

Obk<n O<k$n

CZJ,+3&+3y+(n+l).

Sure enough, the

L&‘S

cancel, and we have enough information to determine

Method 2’:

Cl, without relying on induction:

Perturb your TA.

30,

(n+l)3-3(n+l)n/2-(n+l)

(n+l)(n2+2n+l-3

n-l) =

(n+l)(n+t)n.

Method 3: Build a repertoire.

A slight generalization of the recurrence (2.7) will also suffice for

sum-

mands involving

n2.

The solution to

0~;

R, =

R,P1+(3+yn+6n2,

for n > 0,

(2.4”)

will be of the general form

A(n)ol+B(n)fi + C(n)Y+D(u)d;

(2.41)

and we have already determined A(n), B(n), and C(n), because (2.41) is the

same as (2.7) when 6 = 0. If we now plug in R, =

n3,

we find that

is the

2.5 GENERAL METHODS 45

The horizontal scale

here is ten times the

vertical scale.

solution when a =

-3,

Hence

3D(n)

3C(n) + B(n) =

;

this determines D(n).

We’re interested in the sum Cl,, which equals q -1 + n2; thus we get

17, =

if we set a =

= y = 0 and 6 = 1 in (2.41). Consequently

El, = D(n). We needn’t do the algebra to compute D(n) from B(n) and

C(n), since we already know what the answer will be; but doubters among us

should be reassured to find that

3D(n) =

n3+3C(n)-B(n)

n3+3T-n

n(n+t)(n+I),

Method 4: Replace sums by integrals.

People who have been raised on calculus instead of discrete mathematics

tend to be more familiar with

than with

so they find it natural to try

changing

One of our goals in this book is to become so comfortable

with

that we’ll think s is more difficult than

(at least for exact results).

But still, it’s a good idea to explore the relation between

and J, since

summation and integration are based on very similar ideas.

In calculus, an integral can be regarded as the area under a curve, and we

can approximate this area by adding up the areas of long, skinny rectangles

that touch the curve. We can also go the other way if a collection of long,

skinny rectangles is given: Since Cl, is the sum of the areas of rectangles

whose sizes are 1 x

1 x 4, . . . ,

1 x n2, it is approximately equal to the area

under the curve f(x) = x2 between 0 and n.

f(x

123

The area under this curve is

J,”

x2 dx = n3/3; therefore we know that

El,

approximately

fn3.

46 SUMS

One way to use this fact is to examine the error in the approximation,

in3.

Since q

satisfies the recurrence

[7,-l

+ n2, we find

that

satisfies the simpler recurrence

En =

II,-fn3

IJP1

+n2-in3

E,p1+~(n-1)3+n2-3n3

E,-1

+n-5.

Another way to pursue the integral approach is to find a formula for

summing the areas of the wedge-shaped error terms. We have

x2dx =

(k2-/;P,x2dx)

(k-

1)3

f(k-f)

k=l

Either way, we could find

and then

!I,.

Method 5: Expand and contract.

Yet another way to discover a closed form for Cl, is to replace the orig-

inal sum by a seemingly more complicated double sum that can actually be

simplified if we massage it properly:

(F)(n-j+l)

l<j$n

(n(n+l)+j-j2)

l<j<n

$n2(n+1)+$n(n+1)-50,

tn(n+

t)(n+

,-ton.

Going from a single sum to a double sum may appear at first to be a backward

step, but it’s actually progress, because it produces sums that are easier to

work with. We can’t expect to solve every problem by continually simplifying,

simplifying, and simplifying: You can’t scale the highest mountain peaks by

climbing only uphill!

Method 6: Use finite calculus.

Method 7: Use generating functions.

Stay tuned for still more exciting calculations of

Cl,,

= ,TL=, k2, as we

learn further techniques in the next section and in later chapters.

This is for people

addicted to calculus.

[The last step here

is something like

the last step of

the perturbation

method, because

we get an equation

with the unknown

quantity on both

sides.)

2.6 FINITE AND INFINITE CALCULUS 47

2.6 FINITE AND INFINITE CALCULUS

We’ve learned a variety of ways to deal with sums directly. Now it’s

time to acquire a broader perspective, by looking at the problem of summa-

tion from a higher level. Mathematicians have developed a “finite calculus,”

analogous to the more traditional infinite calculus, by which it’s possible to

approach summation in a nice, systematic fashion.

Infinite calculus is based on the properties of the derivative operator D,

defined by

Df(x) =

:rnO

f(x+

f(x)

’

Finite calculus is based on the properties of the difference operator A, defined

Af(x) = f(x + 1) -f(x).

(2.42)

As opposed to a

cassette function.

This is the finite analog of the derivative in which we restrict ourselves to

positive integer values of h. Thus, h = 1 is the closest we can get to the

“limit” as h

0, and Af(x) is the value of (f(x + h)

f(x))/h when h = 1.

The symbols D and A are called operators because they operate on

functions to give new functions; they are functions of functions that produce

functions. If f is a suitably smooth function of real numbers to real numbers,

then Df is also a function from reals to reals. And if f is any real-to-real

function, so is Af. The values of the functions Df and Af at a point x are

given by the definitions above.

Early on in calculus we learn how D operates on the powers f(x) =

x"'.

In such cases

Df(x) =

mxmP’. We can write this informally with f omitted,

D(xm) =

mx”-‘,

It would be nice if the A operator would produce an equally elegant result;

unfortunately it doesn’t. We have, for example,

A(x3) =

(x+~)~-x’

= 3x

+3x+1.

Math power.

But there is a type of “mth power” that does transform nicely under A,

and this is what makes finite calculus interesting. Such newfangled mth

powers are defined by the rule

factors

= Ix(x-l)...(x-mmlj, integer m 3 0.

(2.43)

Notice the little straight line under the m; this implies that the m factors

are supposed to go down and down, stepwise. There’s also a corresponding

48 SUMS

definition where the factors go up and up:

m factors

iii

x(x+l)...(x+m-l),

integer m 3 0.

(2.44)

When m = 0, we have

= x-

’

= 1, because a product of no factors is

conventionally taken to be 1 (just as a sum of no terms is conventionally 0).

The quantity xm is called “x to the m falling,” if we have to read it

aloud; similarly, xK is “x to the m rising!’ These functions are also called

falling factorial powers and rising factorial powers, since they are closely

related to the factorial function n! = n(n

1). . . (1). In fact, n! =

= 1”.

Several other notations for factorial powers appear in the mathematical

literature, notably “Pochhammer’s symbol” (x), for xK or xm; notations

xc”‘)

or xlml are also seen for x3. But the underline/overline convention

is catching on, because it’s easy to write, easy to remember, and free of

redundant parentheses.

Falling powers xm are especially nice with respect to A. We have

A(G)

(x+1)=-x”

(x+1)x..

.(x-m++)

x...

(x--+2)(x-m+l)

mx(x-l)...(x-m+2),

hence the finite calculus has a handy law to match D(x”‘) = mx”-‘:

A(x”) =

mxd.

(2.45)

This is the basic factorial fact.

The operator D of infinite calculus has an inverse, the anti-derivative

(or integration) operator

The Fundamental Theorem of Calculus relates D

g(x)

Df(xl

if and only if

g(x) dx = f(x) + C.

Here s g(x) dx, the indefinite integral of g(x), is the class of functions whose

derivative is g(x). Analogously, A has as an inverse, the anti-difference (or

summation) operator

and there’s another Fundamental Theorem:

g(x)

Af(xl

if and only if

xg(x)bx

= f(x)+C. (2.46)

Here

g(x) 6x, the indefinite sum of g(x), is the class of functions whose

diflerence is g(x). (Notice that the lowercase

relates to uppercase A as

d relates to D.) The “C” for indefinite integrals is an arbitrary constant; the

“C” for indefinite sums is any function p(x) such that p(x + 1) = p(x). For

Mathematical

terminology is

sometimes crazy:

Pochhammer

12341

actually used the

notation (x)

for the binomial

coefficient

(k)

, not

for factorial powers.

“Quemadmodum

ad differentiam

denotandam

usi

sumus

sign0

ita summam indi-

cabimus sign0

. . .

quo

zquatio

z =

Ay,

siinver-

tatur,

dabit quoque

y = iEz+C.”

-L.

Euler

/88]

You call this a

In other words, the definite sum is the same as an ordinary sum with limits,

punch line?

but excluding the value at the upper limit.

2.6 FINITE AND INFINITE CALCULUS 49

example, C might be the periodic function a + b sin2nx; such functions get

washed out when we take differences, just as constants get washed out when

we take derivatives. At integer values of x, the function C is constant.

Now we’re almost ready for the punch line. Infinite calculus also has

definite integrals: If g(x) = Df(x), then

/‘g(x)dx

f(x)11

= f(b) -f(a).

Therefore finite calculus-ever mimicking its more famous cousin- has def-

inite Sims: If g(x) = Af(x), then

g(x) 6x =

f(x)i’

= f(b) -f(a).

(2.47)

This formula gives a meaning to the notation x.“, g(x) 6x, just as the previous

formula defines

g(x) dx.

But what does

g(x) 6x really mean, intuitively? We’ve defined it by

analogy, not by necessity. We want the analogy to hold, so that we can easily

remember the rules of finite calculus; but the notation will be useless if we

don’t understand its significance. Let’s try to deduce its meaning by looking

first at some special cases, assuming that g(x) = Af(x) = f(x + 1) -f(x). If

b = a, we have

tIg(x)bx

= f(a)-f(a) = 0.

Next, if b = a +

the result is

xl+’

g(x)

f(a+

1) -f(a) = g(a).

More generally, if b increases by

we have

g(x) 6x =

(f(b

+ 1) -f(a))

(f(b) -f(a))

= f(b+ 1) -f(b) = g(b).

These observations, and mathematical induction, allow us to deduce exactly

what x.“, g(x) 6x means in general, when a and b are integers with b > a:

~-$xi~x

~g&,

g(k),

k=a

a<k<b

for integers b 3 a. (2.48)

50 SUMS

Let’s try to recap this; in a slightly different way. Suppose we’ve been

given an unknown sum that’s supposed to be evaluated in closed form, and

suppose we can write it in the form

taskcb

g(k) = I.“, g(x) 6x. The theory

of finite calculus tells us that we can express the answer as f(b)

f(a), if

we can only find an indefinite sum or anti-difference function f such that

g(x) = f (x + 1)

f(x).

C)ne

way to understand this principle is to write

aGk<b

g(k) out in full, using the three-dots notation:

(f(kf1)

-f(k)) =

(f(a+l)

-f(a)) +

(f(a+2)

-f(a+l))

f...

a<k<b

(f(b-1)

f(b-2)) + (f(b)

f(b-1)) .

Everything on the right-ha:nd side cancels, except f(b)

f(a); so f(b)

f(a)

is the value of the sum. (Sums of the form

,Yaskib(f(k

+ 1)

f(k)) are

often called telescoping, by analogy with a collapsed telescope, because the

thickness of a collapsed telescope is determined solely by the outer radius of

And all this time

the outermost tube and the inner radius of the innermost tube.)

I thought it was

But rule (2.48) applies only when b 3 a; what happens if b < a? Well,

telescoping because

collapsed

from

(2.47) says that we mUSt have

very long expression

to a very short one.

g(x) 6x = f(b) -f(a)

= -(f(a)-f(b)) =

-t,“g(x)tx.

This is analogous to the corresponding equation for definite integration. A

similar argument proves

x.‘,,

the summation analog of the iden-

tity

jz.

In full garb,

g(x)

L?(X)

6x,

a a

(2.49)

for all integers a, b, and c.

At this point a few of us are probably starting to wonder what all these

parallels and analogies buy us. Well for one, definite summation gives us a

Others have been

simple way to compute sums of falling powers: The basic laws (2.45),

(2.47),

and (2.48) imply the general law

zify!$

zi,for

nm+’

k”=-

for integers

n 3 0.

(2.50)

O<k<n

m+lo

m+l’

This formula is easy to remember because it’s so much like the familiar

sit

x”’

dx =

n”‘+‘/(m+

1).

2.6 FINITE AND INFINITE CALCULUS 51

In particular, when m = 1 we have

= k, so the principles of finite

calculus give us an easy way to remember the fact that

k =

n(n-1)/2

OS-kin

The definite-sum method also gives us an inkling that sums over the range

0 $ k < n often turn out to be simpler than sums over 1 < k 6 n; the former

are just f(n)

(0))

while the latter must be evaluated as f (n + 1)

f ( 1)

Ordinary powers can also be summed in this new way, if we first express

them in terms of falling powers. For example,

hence

OSk<n

z+:

in(n-l)(n-2+;)

$n(n-i)(n-1).

With friends like

this..

Replacing n by n + 1 gives us yet another way to compute the value of our

old friend q

= ~O~k~n

in closed form.

Gee, that was pretty easy. In fact, it was easier than any of the umpteen

other ways that beat this formula to death in the previous section. So let’s

try to go up a notch, from squares to cubes: A simple calculation shows that

kL+3kL+kL.

(It’s always possible to convert between ordinary powers and factorial powers

by using Stirling numbers, which we will study in Chapter 6.) Thus

Falling powers are therefore very nice for sums. But do they have any

other redeeming features? Must we convert our old friendly ordinary powers

to falling powers before summing, but then convert back before we can do

anything else? Well, no, it’s often possible to work directly with factorial

powers, because they have additional properties. For example, just as we

have (x + y)’ = x2 + 2xy + y2, it turns out that (x +

y)’

= x2 +

2x!-yl+

yz,

and the same analogy holds between (x + y)” and (x +

y)“.

(This “factorial

binomial theorem” is proved in exercise 5.37.)

So far we’ve considered only falling powers that have nonnegative expo-

nents. To extend the analogies with ordinary powers to negative exponents,

52 SUMS

we need an appropriate definition of ~3 for m < 0. Looking at the sequence

x(x-1)(x-2),

= x(x-l),

= x,

= 1,

we notice that to get from

to xl to

we divide by x

2, then

by x

1, then by

It seems reasonable (if not imperative) that we should

divide by x + 1 next, to get from

x5,

thereby making

= 1 /(x + 1).

Continuing, the first few negative-exponent falling powers are

x;1

x+1

x-2

= (x+*:(x+2)

x-3

(x+1)(x+2)(x+3)

and our general definition for negative falling powers is

'-"'

(x+l)(x+2)...(x+m)

for m

(2.51)

(It’s also possible to define falling powers for real or even complex m, but we

How can a

complex

will defer that until Chapter 5.)

number be even?

With this definition, falling powers have additional nice properties. Per-

haps the most important is a general law of exponents, analogous to the law

m+n

XmXn

for ordinary powers. The falling-power version is

xmi-n

xZ(x-m,)n,

integers m and n.

For example,

2)z;

and with a negative n we have

(2.52)

x23

xqx-q-3

= x(x- 1)

(x- 1)x(x+ 1)

x;l,

x+1

If we had chosen to define

as l/x instead of as 1

/(x

l),

the law of

exponents (2.52) would have failed in cases like m = -1 and n = 1. In fact,

we could have used (2.52) to tell us exactly how falling powers ought to be

defined in the case of negative exponents, by setting m = -n. When an

Laws have their

existing notation is being extended to cover more cases, it’s always best to

exponents and their

formulate definitions in such. a way that general laws continue to hold.

detractors.

2.6 FINITE AND INFINITE CALCULUS 53

Now let’s make sure that the crucial difference property holds for our

newly defined falling powers. Does

Ax2

mx*

when m < O? If m = -2,

for example, the difference is

(x+2)(x+3)

(x+1)(x+2)

(x+1)-(x+3)

(x+1)(%+2)(x+3)

-2y-3,

Yes -it works! A similar argument applies for all m < 0.

Therefore the summation property (2.50) holds for negative falling powers

as well as positive ones, as long as no division by zero occurs:

Xmfl

x”&

for mf-1

m+l

(1’

But what about when m =

-l?

Recall that for integration we use

x-’

dx = lnx

when m = -1. We’d like to have a finite analog of lnx; in other words, we

seek a function

f(x)

such that

x-'

= Af(x) = f(x+ 1)-f(x).

x+1

It’s not too hard to see that

f(x) =

;

f...f

;

0.577

exactly?

Maybe they mean

l/d.

Then again,

maybe not.

is such a function, when x is an integer, and this quantity is just the harmonic

number H, of (2.13). Thus H, is the discrete analog of the continuous lnx.

(We will define H, for noninteger x in Chapter 6, but integer values are good

enough for present purposes. We’ll also see in Chapter 9 that, for large x, the

value of H,

In x is approximately 0.577 +

1/(2x).

Hence H, and In x are not

only analogous, their values usually differ by less than 1.)

We can now give a complete description of the sums of falling powers:

ifmf-1;

x”6x

(2.53)

ifm=-1.

54 SUMS

This formula indicates why harmonic numbers tend to pop up in the solutions

to discrete problems like the analysis of quicksort, just as so-called natural

logarithms arise naturally in the solutions to continuous problems.

Now that we’ve found an analog for lnx, let’s see if there’s one for e’.

What function f(x) has the property that Af(x) = f(x), corresponding to the

identity De” = e”? Easy:

f(x+l)-f(X) = f(x)

f(x+ 1) =

2f(x);

so we’re dealing with a simple recurrence, and we can take f(x) = 2” as the

discrete exponential function.

The difference of

is also quite simple, for arbitrary c, namely

A(?)

= cx+’

= (c

1)~“.

Hence the anti-difference of

is c’/(c

if c #

This fact, together with

the fundamental laws (2.47) and (2.48), gives us a tidy way to understand the

general formula for the sum of a geometric progression:

a<k<b

for c # 1.

Every time we encounter a function f that might be useful as a closed

form, we can compute its difference Af = g; then we have a function g whose

indefinite sum

g(x) 6x is known. Table 55 is the beginning of a table of

‘Table 55’ is

difference/anti-difference pairs useful for summation.

page 55. Get it?

Despite all the parallels between continuous and discrete math, some

continuous notions have no discrete analog. For example, the chain rule of

infinite calculus is a handy rule for the derivative of a function of a function;

but there’s no corresponding chain rule of finite calculus, because there’s no

nice form for Af (g (x)) . Discrete change-of-variables is hard, except in certain

cases like the replacement of x by c

However, A(f(x) g(x))

oes

have a fairly nice form, and it provides us

with a rule for summation by parts, the finite analog of what infinite calculus

calls integration by parts. Let’s recall that the formula

D(uv) =

uDv+vDu

of infinite calculus leads to

t’he

rule for integration by parts,

uDv

= uv-

VDU,

Infinite calculus

avoids E

here by

letting

1 -3 0.

guess

2”)

for

small values of

2.6 FINITE AND INFINITE CALCULUS 55

Table 55 What’s the difference?

f =

Af = g

x2=x(x-l)

mxti

xmf'/(m+l)

x-‘=

l/(x+1)

f=Lg

Af = g

)cX

c"/(c-1)

cAf

f+g

Af+Ag

fAg

+ EgAf

after integration and rearranging terms; we can do a similar thing in finite

calculus.

We start by applying the difference operator to the product of two func-

tions u(x) and v(x):

A@(x) v(x)) =

u(x+l)

v(x+l)

u(x) v(x)

= u(x+l)v(x+l)-u(x)v(x+l)

+u(x)v(x+l)-u(x)v(x)

= u(x) Av(x) +

v(x+l)

Au(x).

(2.54)

This formula can be put into a convenient form using the

shij?!

operator E,

defined by

Ef(x) = f(x+l).

Substituting this for

v(x+l)

yields a compact rule for the difference of a

product:

A(uv)

= uAv + EvAu.

(2.55)

(The E is a bit of a nuisance, but it makes the equation correct.) Taking

the indefinite sum on both sides of this equation, and rearranging its terms,

yields the advertised rule for summation by parts:

uAv =

uv-

EvAu.

(2.56)

As with infinite calculus, limits can be placed on all three terms, making the

indefinite sums definite.

This rule is useful when the sum on the left is harder to evaluate than the

one on the right. Let’s look at an example. The function s xe’ dx is typically

integrated by parts; its discrete analog is

x2’

6x, which we encountered

earlier this chapter in the form

xt=,

k2k. To sum this by parts, we let

56 SUMS

u(x) = x and

Av(x)

2’;

hence Au(x) = 1, v(x) =

2x,

and

Ev(x)

2X+1.

Plugging into (2.56) gives

x2”

= x2”

2X+’

6x = x2” -

2x+’

And we can use this to evaluate the sum we did before, by attaching limits:

k2k

t;+‘x2”

k=@

x2X-2X+l

ll+’

= ((n-t

1)2”+’

-2n+2)

(0.2’-2’)

= (n-

1)2n+’

f2.

It’s easier to find the sum this way than to use the perturbation method,

because we don’t have to tlrink.

The ultimate

goal

We stumbled across a formula for toSk<,,

earlier in this chapter,

!fmat!ernatics

and counted ourselves lucky. But we could have found our formula (2.36)

systematically, if we had known about summation by parts. Let’s demonstrate

~~~$~/~t$$rt

thought.

this assertion by tackling a sum that looks even harder, toSk<,,

kHk.

The

solution is not difficult if we are guided by analogy with

x dx: We take

u(x) =

and

Av(x)

= x

x1,

hence Au(x) =

x5,

v(x) =

x2/2,

Ev(x)

(x +

1)2/2,

and we have

1)’

xxH,Sx

;Hx

x-’

;Hx

fxx16x

(In going from the first line to the second, we’ve combined two falling pow-

ers

(x+1)2x5

by using the law of exponents (2.52) with m = -1 and n = 2.)

Now we can attach limits and conclude that

kHk =

t;xHx6x

;(Hn-;),

OSk<n

2.7 INFINITE SUMS

(2.57)

When we defined t-notation at the beginning of this chapter, we

finessed the question of infinite sums by saying, in essence, “Wait until later.

is finesse?

For now, we can assume that all the sums we meet have only finitely many

nonzero terms.”

But the time of reckoning has finally arrived; we must face

Sure: 1 + 2 +

4 + 8 + . . is the

“infinite precision”

representation of

the number -1,

in a binary com-

puter with infinite

word size.

2.7 INFINITE SUMS 57

the fact that sums can be infinite. And the truth is that infinite sums are

bearers of both good news and bad news.

First, the bad news: It turns out that the methods we’ve used for manip-

ulating

1’s

are not always valid when infinite sums are involved. But next,

the good news: There is a large, easily understood class of infinite sums for

which all the operations we’ve been performing are perfectly legitimate. The

reasons underlying both these news items will be clear after we have looked

more closely at the underlying meaning of summation.

Everybody knows what a finite sum is: We add up a bunch of terms, one

by one, until they’ve all been added. But an infinite sum needs to be defined

more carefully, lest we get into paradoxical situations.

For example, it seems natural to define things so that the infinite sum

s =

l+;+;+f+&+&+...

is equal to 2, because if we double it we get

2s =

2+1+;+$+;+$+.-

2+s.

On the other hand, this same reasoning suggests that we ought to define

T =

1+2+4+8+16+32-t...

to be -1, for if we double it we get

2T =

2+4+8+16+32+64+...

= T-l.

Something funny is going on; how can we get a negative number by summing

positive quantities? It seems better to leave T undefined; or perhaps we should

say that T = 00, since the terms being added in T become larger than any

fixed, finite number. (Notice that

is another “solution” to the equation

2T = T

1; it also “solves” the equation 2S = 2 + S.)

Let’s try to formulate a good definition for the value of a general sum

kEK

ok, where K might be infinite. For starters, let’s assume that all the

terms

are nonnegative. Then a suitable definition is not hard to find: If

there’s a bounding constant A such that

for all finite subsets F c K, then we define

tkeK

to be the least such A.

(It follows from well-known properties of the real numbers that the set of

all such A always contains a smallest element.) But if there’s no bounding

constant A, we say that ,YkEK

= 00; this means that if A is any real

number, there’s a set of finitely many terms

whose sum exceeds A.

58 SUMS

The definition in the previous paragraph has been formulated carefully

so that it doesn’t depend on any order that might exist in the index set K.

Therefore the arguments we are about to make will apply to multiple sums

with many indices kl , k2, . . ,

not just to sums over the set of integers.

In the special case that K is the set of nonnegative integers, our definition

for nonnegative terms

implies that

Here’s why: Any nondecreasing sequence of real numbers has a limit (possi-

bly

ok).

If the limit is A, and if F is any finite set of nonnegative integers

whose elements are all 6 n, we have

tkEF

ok 6 ~~Zo

< A; hence A =

or A is a bounding constant. And if A’ is any number less than the stated

limit A, then there’s an n such that

~~=,

> A’; hence the finite set

={O,l,...

,n} witnesses to the fact that A’ is not a bounding constant.

We can now easily

com,pute

the value of certain infinite sums, according

to the definition just given. For example, if

= xk, we have

The set K might

even be uncount-

able. But only a

countable num-

ber of terms can

be nonzero, if a

bounding constant

A exists, because at

most nA terms are

3 l/n.

In particular, the infinite sums S and T considered a minute ago have the re-

spective values 2 and

co,

just as we suspected. Another interesting example is

l.im~k~=J~m~_l

=l.

n-+cc

k=O

Now let’s consider the

‘case

that the sum might have negative terms as

well as nonnegative ones. What, for example, should be the value of

E(-1)k =

l-l+l--l+l-l+~~~?

k>O

If we group the terms in pairs, we get

“Aggregatum

quantitatum

a-a+a-a+a--a

etc.

nunc

est = a,

(l--1)+(1-1)+(1-1)+... =

O+O+O+...

)

nunc

= 0, adeoque

continuata in

infini-

so the sum comes out zero; but if we start the pairing one step later, we get

turn serie ponendus

= a/2,

fateor

‘-(‘-‘)-(1-1)-(1-l)-...

‘-O-O-O-...;

acumen et veritatem

animadversionis

ture.”

-G.

Grandi

1133)

the sum is

2.7 INFINITE SUMS 59

We might also try setting x = -1 in the formula

xk = 1

/(l

x),

since we’ve proved that this formula holds when 0 < x < 1; but then we are

forced to conclude that the infinite sum is

although it’s a sum of integers!

Another interesting example is the doubly infinite

where

l/(k+

1) for k 3 0 and

l/(k-

1) for k < 0. We can write this as

.'.+(-$)+(-f)+(-;)+l+;+f+;+'.'.

(2.58)

If we evaluate this sum by starting at the “center” element and working

outward,

..+

(-$+(-f

+(-;

+(l)+

;,+

g-t

;> +...,

we get the value 1; and we obtain the same value 1 if we shift all the paren-

theses one step to the left,

+(-j+(-;+cf+i-;)+l)+;)+:)+.y

because the sum of all numbers inside the innermost n parentheses is

-----...-

j+,+;+...+L

l-L_

nfl

n n-l n

K-3’

A similar argument shows that the value is 1 if these parentheses are shifted

any fixed amount to the left or right; this encourages us to believe that the

sum is indeed

On the other hand, if we group terms in the following way,

..+(-i+(-f+(-;+l+;,+f+;)+;+;)+...,

the nth pair of parentheses from inside out contains the numbers

----

-...-

2+,+;+...+

n+l n

+ Hz,,

&+I

We’ll prove in Chapter 9 that

lim,,,

(Hz,-H,+,

) = ln2; hence this grouping

suggests that the doubly infinite sum should really be equal to 1 + ln2.

There’s something flaky about a sum that gives different values when

its terms are added up in different ways.

Advanced texts on analysis have

a variety of definitions by which meaningful values can be assigned to such

pathological sums; but if we adopt those definitions, we cannot operate with

x-notation as freely as we have been doing. We don’t need the delicate refine-

ments of “conditional convergence”

for the purposes of this book; therefore

Is this the first page

we’ll stick to a definition of infinite sums that preserves the validity of all the

with no graffiti?

operations we’ve been doing in this chapter.

60 SUMS

In fact, our definition of infinite sums is quite simple. Let K be any

set, and let

be a real-valued term defined for each k

K. (Here ‘k’

might actually stand for several indices kl , k2, . . , and K might therefore be

multidimensional.) Any real number x can be written as the difference of its

positive and negative parts,

x+-x

where x+

=x.[x>O]

and

-x.[x<Ol.

(Either x+=Oorx

= 0.) We’ve already explained how to define values for

the infinite sums

kEK

‘:

and

tkEK

because

and a{ are nonnegative.

Therefore our general definition is

(2.59)

kEK

kGK

unless the right-hand sums are both equal to

co.

In the latter case, we leave

keK

undefined.

Let A+ =

,YkEK

and A- =

tktK

ai.

If A+ and A- are both finite,

the sum

tkEK

ok is said to converge absolutely to the value A =

A+

A-.

In other words,

ab-

A+

but A is finite, the sum

tkeK

is said to diverge to

+a.

so1ute

convergence

Similarly, if A- =

but A+ is finite,

tktK

is said to diverge to

--oo.

$e~~~o~o:,“,a,“~~~U~~m

A+

= A- = 00, all bets are off.

converges.

We started with a definition that worked for nonnegative terms, then we

extended it to real-valued terms. If the terms

are complex numbers, we

can extend the definition

on.ce

again, in the obvious way: The sum tkeK

is defined to be tkCK

%ok

itk,-K

Jok, where 3iok and 3ok are the real

and imaginary parts of ok--provided that both of those sums are defined.

Otherwise

tkEk

is undefined. (See exercise 18.)

The bad news, as stated earlier, is that some infinite sums must be left

undefined, because the manipulations we’ve been doing can produce inconsis-

tencies in all such cases. (See exercise 34.) The good news is that all of the

manipulations of this chapter are perfectly valid whenever we’re dealing with

sums that converge absolutely, as just defined.

We can verify the good news by showing that each of our transformation

rules preserves the value of all absolutely convergent sums. This means, more

explicitly, that we must prove the distributive, associative, and commutative

laws, plus the rule for summing first on one index variable; everything else

we’ve done has been derived from those four basic operations on sums.

The distributive law (2.15) can be formulated more precisely as follows:

tkEK

converges absolmely to A and if c is any complex number, then

keK

cok

converges absolutely to

CA.

We can prove this by breaking the sum

into real and imaginary, positive and negative parts as above, and by proving

the special case in which c

0 and each term

is nonnegative. The proof

2.7 INFINITE SUMS 61

in this special case works because tkEF

cok

= c

tkeF

for all finite

Sets

the latter fact follows by induction on the size of F.

The associative law (2.16) can be stated as follows: If tkEK

and

tkeK

converge absolutely to A and B, respectively, then

tkek(ok

bk)

converges absolutely to A + B. This turns out to be a special case of a more

general theorem that we will prove shortly.

The commutative law (2.17) doesn’t really need to be proved, because

we have shown in the discussion following (2.35) how to derive it as a special

case of a general rule for interchanging the order of summation.

The main result we need to prove is the fundamental principle of multiple

sums: Absolutely convergent sums over two or more indices can always be

summed first with respect to any one of those indices. Formally, we shall

Best to skim this

prove that if J and the elements of

{Ki

1 j

J} are any sets of indices such that

page the first time

you get here.

Your friendly

oi,k

converges absolutely to A,

iEJ

kEKj

then there exist complex numbers

for each j

J such that

oj,k

converges absolutely to Aj, and

&K,

Aj converges absolutely to A.

iEJ

It suffices to prove this assertion when all terms are nonnegative, because we

can prove the general case by breaking everything into real and imaginary,

positive and negative parts as before. Let’s assume therefore that

oi,k

3 0 for

all pairs (j, k)

M, where M is the master index set {(j, k) 1 j

J, k

Kj}.

We are given that

tCj,k)EM

oj,k

is finite, namely that

aj,k

(j.k)EF

for all finite subsets F

M, and that A is the least such upper bound. If j is

any element of J, each sum of the form

xkEFi

oj,k

where

is a finite subset

is bounded above by A. Hence these finite sums have a least upper

bound Ai 3 0, and tkEKi

oj,k

= Aj by definition.

We still need to prove that A is the least upper bound of xjEG Aj,

for all finite subsets G

J. Suppose that G is a finite subset of J with

xjEG Aj = A’ > A. We CXI find finite subsets

such that tkeFi

oj,k

(A/A’)Aj for each j

G with Aj > 0. There is at least one such j. But then

~.iEG,kCFi

oj,k

> (A/A’) xjEG Aj = A, contradicting the fact that we have

62 SUMS

tCj,kiEF

< A for all finite subsets F

M. Hence

xjEG

< A, for all

finite subsets G

Finally, let A’ be any real number less than A. Our proof will be complete

if we can find a finite set G

J such that

xjeo

> A’. We know that

there’s a finite set F

M such that &j,kIeF

oj,k

> A’; let G be the set of j’s

in this F, and let

= {k 1 (j, k)

F}. Then

xjeG

xjEG

tkcF,

oj,k

t(j,k)EF aj,k > A’;

QED.

OK, we’re now legitimate! Everything we’ve been doing with infinite

sums is justified, as long a3 there’s a finite bound on all finite sums of the

absolute values of the terms. Since the doubly infinite sum (2.58) gave us

two different answers when we evaluated it in two different ways, its positive

whY

have

been

terms 1 +

. . must diverge to 03; otherwise we would have gotten the

hearing a lot lately

about “harmonic

same answer no matter how we grouped the terms.

convergence”?

Exercises

Warmups

What does the notation

k=4

mean?

Simplify the expression x . ([x >

< 01).

3 Demonstrate your understanding of t-notation by writing out the sums

in full. (Watch out -the second sum is a bit tricky.)

4 Express the triple sum

aijk

lSi<j<k<4

as a three-fold summation (with three

x’s),

summing first on k, then j, then i;

summing first on i, then j, then k.

Also write your triple sums out in full without the t-notation, using

parentheses to show what is being added together first.

2 EXERCISES 63

5 What’s wrong with the following derivation?

6 What is the value of

tk[l

6 j $

n], as a function of j and n?

Yield to the rising

7 Let Vf(x) = f(x)

f(x-1). What is

V(xm)?

power.

8 What is the value of O”, when m is a given integer?

9 What is the law of exponents for rising factorial powers, analogous to

(2.52)? Use this to define

XC”.

The text derives the following formula for the difference of a product:

A(uv)

= uAv + EvAu.

How can this formula be correct, when the left-hand side is symmetric

with respect to u and v but the right-hand side is not?

Basics

The general rule (2.56) for summation by parts is equivalent to

ak+l

ak)bk

anbn

aOb0

O$k<n

-t

%+I

h+l

bd, for

O<k<n

Prove this formula directly by using the distributive, associative, and

commutative laws.

Show that the function p(k) =

(-l)k~

is a permutation of the set of

all integers, whenever c is an integer.

13 Use the repertoire method to find a closed form for

xr=o(-l)kk2.

14 Evaluate

xi=,

k2k by rewriting it as the multiple sum

tlbjGkGn

2k.

15 Evaluate

Gil,,

EL=,

by the text’s Method 5 as follows: First write

+ q n =

xl$j<k$n

jk;

then

aPPlY

(2.33).

16 Prove that

x”/(x

n)”

x3/(x

m)n, unless one of the denominators

is zero.

17 Show that the following formulas can be used to convert between rising

and falling factorial powers, for all integers m:

iii

X = (-l)"(-x)2 =

(x+m-1)"

l/(x-l)=;

xl'l.

(-l)"(-x)"

(x-m+l)"

= l/(x+1)-m.

(The answer to exercise 9 defines x-“‘.)

64 SUMS

18 Let

9%~

and Jz be the real and imaginary parts of the complex num-

ber

The absolute value

Iz/

J(!??z)~

(3~)~.

A sum

tkeK

of com-

plex terms

is said to converge absolutely when the real-valued sums

t&K

*ak

and

tkEK

?ok both converge absolutely. Prove that tkEK

converges absolutely if and only if there is a bounding constant B such

that xkEF

[oki

< B for

,a11

finite subsets F

Homework exercises

Use a summation factor to solve the recurrence

= 5;

2T,,

nT,-,

+ 3 . n! ,

for n > 0.

Try to evaluate ~~=, kHk by the perturbation method, but deduce the

VdUe

~~=:=,

instead.

Evaluate the sums S, =

xc=o(-l)n-k,

= ~~=o(-l)n-kk, and Ll, =

t;=o(-l)n-kk2

the perturbation method, assuming that n 3 0.

Prove Lagrange’s identity (without using induction):

It’s hard to prove

the

identity

(Cljbk-Clkbj)2 =

(~Cl~)(~b~)

(LClkbk)‘.

<j<k<n

k=l k=l

This, incidentally, implies Cauchy’s inequality,

akbb)l

bZk)

k:=l

k=l

Evaluate the sum

Et=:=,

(2k + 1 )/(k(k + 1)) in two ways:

Replace 1

/k(k

1) by the “partial fractions” 1

/(k

+ 1).

b Sum by parts.

What is to<k<n

&/(k

+ l)(k +

2)?

Hint: Generalize the derivation of

(2.57).

The notation

nk,k

ok means the product of the numbers ok for all k

K. This

notation was

Assume for simplicity that

# 1 for only finitely many k; hence infinite

introduced

products need not be defined. What laws does this n-notation satisfy,

Jacobi in 1829

[162].

analogous to the distributive, associative, and commutative laws that

hold for

Express the double product nlsjQkbn

ok in terms of the single product

nEz,

by manipulating n-notation. (This exercise gives us a product

analog

of the upper-triangle identity

(2.33).)

2 EXERCISES 65

27 Compute A(cx), and use it to deduce the value of

xE=,

(-2)k/k.

28 At what point does the following derivation go astray?

==(

k>l

j31

F[j=k+l]-k[j=k-1]

j>l

k>l

;[j=k+l]-k[j=k-1]

)

;[k=j-l]-i[k=j+l]

Exam problems

=x(

j-l j

---

i31

j+l

-'.

29 Evaluate the sum

(-l)kk/(4k2

1).

30 Cribbage players have long been aware that 15 = 7 + 8 = 4 + 5 + 6 =

1 + 2 + 3 + 4 + 5. Find the number of ways to represent 1050 as a sum of

consecutive positive integers. (The trivial representation ‘1050’ by itself

counts as one way; thus there are four, not three, ways to represent 15

as a sum of consecutive positive integers. Incidentally, a knowledge of

cribbage rules is of no use in this problem.)

31 Riemann’s zeta function c(k) is defined to be the infinite sum

Prove that

tka2(L(k)

1) = 1. What is the value of

tk?l

(L(2k)

l)?

Let a 2 b = max(0, a

b). Prove that

tmin(k,x’k)

x(x:

(2k+

1))

k>O

k?O

for all real

3 0, and evaluate the sums in closed form.

Bonus problems

The laws of the

jungle.

Let

/\kcK

denote the minimum of the numbers

(or their greatest

lower bound, if K is infinite), assuming that each

is either real or

foe.

What laws are valid for A-notation, analogous to those that work for

and

(See exercise 25.)

SUMS

Prove that if the sum

tkeK

is undefined according to (zsg), then it

is extremely flaky in the following sense: If

A-

and A+ are any given

real numbers, it’s possible to find a sequence of finite subsets

c Fl c

’

. . of K such that

6 A-,

when n is odd;

A+,

when

is even.

&Fn

kEFn

Prove Goldbach’s theorem

1 =

;+;+;+:;+;+&+$+&+...

t’,

kEP

k-’

where

is the set of “perfect powers” defined recursively as follows:

Perfect

power

corrupts perfectly.

P = {mn 1 m 3 2,n 3 2,m

P}.

Solomon

Golomb’s

“self.-describing sequence” (f (1) , f

(2))

(3))

. . . ) is the

only nondecreasing sequence of positive integers with the property that

it contains exactly f(k) occurrences of k for each k. A few moments’

thought reveals that the sequence must begin as follows:

c+++x:i::::lk2

Let g(n) be the largest integer m such that f(m) = n. Show that

a s(n) = EC=,

f(k).

b 9(9(n)) = Ed=, Wk).

9(9(9(n))) = ing(fl)(g(n) +

IL;::

g(k)(g(k) +

1).

Research

problem

37 Will all the l/k by

l/(k

+ 1) rectangles, for k 3 1, fit together inside a

1 by 1 square? (Recall that their areas sum to

1.1

Integer Functions

)Ouch.(

WHOLE NUMBERS constitute the backbone of discrete mathematics, and we

often need to convert from fractions or arbitrary real numbers to integers. Our

goal in this chapter is to gain familiarity and fluency with such conversions

and to learn some of their remarkable properties.

3.1 FLOORS AND CEILINGS

We start by covering the floor (greatest integer) and ceiling (least

integer) functions, which are defined for all real x as follows:

1x1

= the greatest integer less than or equal to x;

[xl

= the least integer greater than or equal to x .

(3.1)

Kenneth E. Iverson introduced this notation, as well as the names “floor” and

“ceiling,” early in the 1960s

[161,

page

121.

He found that typesetters could

handle the symbols by shaving the tops and bottoms off of

’

[’

and

‘I

‘.

His

notation has become sufficiently popular that floor and ceiling brackets can

now be used in a technical paper without an explanation of what they mean.

Until recently, people had most often been writing

‘[xl’

for the greatest integer

6 x, without a good equivalent for the least integer function. Some authors

had even tried to use

‘]x[‘-with

a predictable lack of success.

Besides variations in notation, there are variations in the functions them-

selves. For example, some pocket calculators have an INT function, defined

1x1

when x is positive and

[xl

when x is negative. The designers of

these calculators probably wanted their INT function to satisfy the iden-

tity INT(-x) =

-INT(x).

But we’ll stick to our floor and ceiling functions,

because they have even nicer properties than this.

One good way to become familiar with the floor and ceiling functions

is to understand their graphs, which form staircase-like patterns above and

68 INTEGER FUNCTIONS

below the line f(x) = x:

We see from the graph that., for example,

lel

l-ej

=-3,

Tel

r-e] = -2,

since e

2.71828.. . .

By staring at this illustration we can observe several facts about floors

and ceilings. First, since the floor function lies on or below the diagonal line

f(x) = x, we have

1x1

similarly [xl 3 x. (This, of course, is quite

obvious from the definition.) The two functions are equal precisely at the

integer points:

lx] =

x is an integer

[xl

= x.

(We use the notation

‘H’

to mean “if and only if!‘) Furthermore, when

they differ the ceiling is exactly 1 higher than the floor:

[xl

1x1

is not an integer] .

(3.2)

Cute.

Iverson

‘s

bracket

If we shift the diagonal line down one unit, it lies completely below the floor

conventions

this

function, so x

1 <

1x1;

similarly x + 1 >

[xl.

Combining these observations

complete equation.

gives us

x-l <

lx]

[xl

x+1.

(3.3)

Finally, the functions are reflections of each other about both axes:

l-XJ

-[xl

;

r-x.1

= -1xJ

(3.4)

3.1 FLOORS AND CEILINGS 69

Next week we’re

getting

walls.

Thus each is easily expressible in terms of the other. This fact helps to

explain why the ceiling function once had no notation of its own. But we

see ceilings often enough to warrant giving them special symbols, just as we

have adopted special notations for rising powers as well as falling powers.

Mathematicians have long had both sine and cosine, tangent and cotangent,

secant and cosecant, max and min; now we also have both floor and ceiling.

To actually prove properties about the floor and ceiling functions, rather

than just to observe such facts graphically, the following four rules are espe-

cially useful:

1x1

n<x<n+l,

(a)

LxJ=n

x-l<n<x,

(b)

[xl=n

n-l

<x<n,

(c)

(3.5)

[xl=n

x$n<x+l.

(We assume in all four cases that n is an integer and that x is real.) Rules

(a) and (c) are immediate consequences of definition

(3.1);

rules (b) and (d)

are the same but with the inequalities rearranged so that n is in the middle.

It’s possible to move an integer term in or out of a floor (or ceiling):

lx + n] =

1x1

+ n,

integer n.

(3.6)

(Because rule (3.5(a)) says that this assertion is equivalent to the inequalities

1x1

+ n < x + n <

Lx]

+ n + 1.) But similar operations, like moving out a

constant factor, cannot be done in general. For example, we have

[nx]

# n[x]

when n = 2 and x =

l/2.

This means that floor and ceiling brackets are

comparatively inflexible. We are usually happy if we can get rid of them or if

we can prove anything at all when they are present.

It turns out that there are many situations in which floor and ceiling

brackets are redundant, so that we can insert or delete them at will. For

example, any inequality between a real and an integer is equivalent to a floor

or ceiling inequality between integers:

x<n

Lx]<n,

n<x

n <

[xl,

(b)

x6n

[xl

Cc)

(3.7)

n6x

n 6

1x1

These rules are easily proved. For example, if x < n then surely

1x1

< n, since

1x1

6 x. Conversely, if

1x1

< n then we must have x < n, since x <

lx]

+ 1

and

1x1

+ 1 < n.

It would be nice if the four rules in (3.7) were as easy to remember as

they are to prove. Each inequality without floor or ceiling corresponds to the

70 INTEGER FUNCTIONS

same inequality with floor or with ceiling; but we need to think twice before

deciding which of the two is appropriate.

The difference between. x and

1x1

is called the fractional part of x, and

it arises often enough in applications to deserve its own notation:

{x}

lx]

(3.8)

We sometimes call

Lx]

the integer part of x, since

1x1

+ {x}. If a real

number x can be written in the form x = n + 8, where n is an integer and

0 <

we can conclude by (3.5(a)) that n =

1x1

and

= {x}.

Identity (3.6) doesn’t hold if n is an arbitrary real. But we can deduce

that there are only two possibilities for lx +

in general: If we write x =

1x1

{x}

and y =

[yJ

{y},

then we have lx +

1x1

LyJ

1(x>

{y}J.

And since 0 < {x} + {y} < 2, we find that sometimes lx +

1x1

+ [y],

otherwise it’s

1x1

[y]

3.2

FLOOR/CEILING APPLICATIONS

We’ve now seen the basic tools for handling floors and ceilings. Let’s

put them to use, starting with an easy problem: What’s [lg351? (We use ‘lg’

to denote the base-2 logarithm.) Well, since

< 35 6 26, we can take logs

to get 5 < lg35 6 6; so (3.5(c)) tells us that [lg35] = 6.

Note that the number 35 is six bits long when written in radix 2 notation:

35 = (100011)~. Is it always true that [lgnl is the length of n written in

binary? Not quite. We also need six bits to write 32 = (100000)2. So [lgnl

is the wrong answer to the problem. (It fails only when n is a power of 2,

but that’s infinitely many failures.) We can find a correct answer by realizing

that it takes m bits to write each number n such that

2”-’

6 n < 2m; thus

&(a))

tells us that m

1 = LlgnJ, so m =

1lgn.J

+ 1. That is, we need

\lgnJ t 1 bits to express n in binary, for all n > 0. Alternatively, a similar

derivation yields the answer [lg(n t 1 )I; this formula holds for n = 0 as well,

if we’re willing to say that it takes zero bits to write n = 0 in binary.

Let’s look next at expressions with several floors or ceilings. What is

[lxJl?

Easy-smce

1x1

is an integer,

[lx]]

is just

1x1.

So is any other ex-

pression with an innermost

1x1

surrounded by any number of floors or ceilings.

Here’s a tougher problem: Prove or disprove the assertion

[JI;TII =

lJ;;I,

real x 3 0.

(3.9)

Equality obviously holds

wh.en

x is an integer, because x =

1x1.

And there’s

equality in the special cases

= 3.14159. . . , e = 2.71828. . . , and

+&)/2

1.61803...,

because we get 1 = 1. Our failure to find a

coun-

terexample suggests that equality holds in general, so let’s try to prove it.

Hmmm. We’d bet-

ter not write {x}

for the fractional

part when it could

be confused with

the set containing x

as its only element.

The second case

occurs if and only

if there’s a “carry”

at the position of

the decimal point,

when the fractional

parts {x} and {y}

are added together.

[Of course

7-c,

and 4 are the

obvious first real

numbers to try,

aren’t they?)

3.2 FLOOR/CEILING APPLICATIONS 71

Skepticism is

healthy only to

a limited extent.

Being skeptical

about proofs and

programs (particu-

larly your own) will

probably keep your

grades healthy and

your job fairly se-

cure. But applying

that much skepti-

cism will probably

also keep you shut

away working all

the

time, instead

of letting you get

out

for

exercise and

relaxation.

Too

much skepti-

cism is an open in-

vitation

to the

state

of rigor mortis,

where you become

so worried about

being correct and

rigorous that you

never get anything

finished.

-A skeptic

(This observation

was made by R. J.

McEliece when he

was an undergrad.)

Incidentally, when we’re faced with a “prove or disprove,” we’re usually

better off trying first to disprove with a counterexample, for two reasons:

A disproof is potentially easier (we need just one counterexample); and nit-

picking arouses our creative juices.

Even if the given assertion is true, our

search for a counterexample often leads us to a proof, as soon as we see why

a counterexample is impossible. Besides, it’s healthy to be skeptical.

If we try to prove that

[m]

= L&J with the help of calculus, we might

start by decomposing x into its integer and fractional parts [xJ + {x} = n +

and then expanding the square root using the binomial theorem: (n+(3)‘/’ =

n’/2 +

n-‘/2(j/2

&/2@/g

+ . . . .

But this approach gets pretty messy.

It’s much easier to use the tools we’ve developed. Here’s a possible strat-

egy: Somehow strip off the outer floor and square root of

[ml,

then re-

move the inner floor, then add back the outer stuff to get

Lfi].

OK. We let

m=llmj

invoke (3.5(a)), giving m 6

< m + 1. That removes

the outer floor bracket without losing any information. Squaring, since all

three expressions are nonnegative, we have m2 6

Lx]

< (m + 1)‘. That gets

rid of the square root. Next we remove the floor, using (3.7(d)) for the left

inequality and (3.7(a)) for the right: m2 6 x < (m +

1)2.

It’s now a simple

matter to retrace our steps, taking square roots to get m 6

< m + 1 and

invoking (3.5(a)) to get m =

[J;;].

Thus

\m]

= m =

l&J;

the assertion

is true. Similarly, we can prove that

[ml

[J;;]

real x 3 0.

The proof we just found doesn’t rely heavily on the properties of square

roots. A closer look shows that we can generalize the ideas and prove much

more: Let f(x) be any continuous, monotonically increasing function with the

property that

f(x) = integer

===3

x = integer.

(The symbol

‘==+I

means “implies!‘) Then we have

lf(x)J

lf(lxJ

and

If(x)1 = Tf(Txl)l,

(3.10)

whenever f(x),

f(lxJ),

and f(

[xl)

are defined. Let’s prove this general prop-

erty for ceilings, since we did floors earlier and since the proof for floors is

almost the same. If x =

[xl,

there’s nothing to prove. Otherwise x <

[xl,

and f(x) < f (

[xl

) since f is increasing. Hence

(x)1

(

[xl

)I,

since

nondecreasing. If

[f(x)]

[f(

[xl)],

there must be a number y such that

[xl

and

f(y)

Tf(x)l,

since f is continuous. This y is an integer, be-

cause of f's special property. But there cannot be an integer strictly between

x and

[xl.

This contradiction implies that we must have

(x)1

(

[xl

)I.

72 INTEGER FUNCTIONS

An important special case of this theorem is worth noting explicitly:

if m and n are integers and the denominator n is positive. For example, let

m = 0; we have

[l[x/lO]/lOJ

/lOI =

[x/1000].

Dividing thrice by 10 and

throwing off digits is the same as dividing by 1000 and tossing the remainder.

Let’s try now to prove or disprove another statement:

This works when x =

and x = e, but it fails when x = 4; so we know that

it isn’t true in general.

Before going any further, let’s digress a minute to discuss different

“lev-

els” of questions that can be asked in books about mathematics:

Level 1. Given an explicit object x and an explicit property P(x), prove that

P(x) is true. For example, “Prove that

1x1

= 3.” Here the problem involves

finding a proof of some purported fact.

Level 2. Given an explicit set X and an explicit property P(x), prove that

P(x) is true for

all

X. For example, “Prove that

1x1

< x for all real x.”

Again the problem involves finding a proof, but the proof this time must be

general. We’re doing algebra, not just arithmetic.

Level 3. Given an explicit set X and an explicit property P(x), prove or

disprove that P(x) is true for all x

X. For example, “Prove or disprove

In my other texts

that

[ml

= [J;;] for all real x 2

0.”

Here there’s an additional level ~~se~~~~nr($

of uncertainty; the outcome might go either way. This is closer to the real

Same

~~~~~~~~~

situation a mathematician constantly faces: Assertions that get into books

about 99.44% df

tend to be true, but new things have to be looked at with a jaundiced eye. If

the

time;

but

not

the statement is false, our job is to find a counterexample. If the statement

in this book.

is true, we must find a proof as in level 2.

Level 4. Given an explicit set X and an explicit property P(x), find a neces-

sary and

suficient

condition Q(x) that P(x) is true. For example, “Find a

necessary and sufficient condition that

1x1

[xl

.”

The problem is to find Q

such that P(x)

Q(x). Of course, there’s always a trivial answer; we can

take Q(x) = P(x). But the implied requirement is to find a condition that’s as

simple as possible. Creativity is required to discover a simple condition that

But

no simpler.

will work. (For example, in this case, “lx] 3

[xl

x is an integer.“) The

-A. Einstein

extra element of discovery needed to find Q(x) makes this sort of problem

more difficult, but it’s more typical of what mathematicians must do in the

“real world!’ Finally, of course, a proof must be given that P(x) is true if and

only if Q(x) is true.

3.2 FLOOR/CEILING APPLICATIONS 73

Level 5. Given an explicit set X, find an interesting property P(x) of its

elements. Now we’re in the scary domain of pure research, where students

might think that total chaos reigns. This is real mathematics. Authors of

textbooks rarely dare to ask level 5 questions.

Home of the

Toledo

Mudhens.

End of digression. But let’s convert our last question from level 3 to

level 4: What is a necessary and sufficient condition that [JLT;Jl =

[fil?

We have observed that equality holds when

= 3.142 but not when x = 1.618;

further experimentation shows that it fails also when

is between 9 and 10.

Oho. Yes. We see that bad cases occur whenever m2 < x < m2 +

since this

gives m on the left and m + 1 on the right. In all other cases where

J;;

defined, namely when x = 0 or m2 + 1 6 x 6 (m + 1

)2,

we get equality. The

following statement is therefore necessary and sufficient for equality: Either

x is an integer or

isn’t.

(Or, by pessimists,

half-closed.)

For our next problem let’s consider a handy new notation, suggested

by C. A. R. Hoare and Lyle Ramshaw, for intervals of the real line:

[01.

denotes the set of real numbers x such that

< x 6

(3.

This set is called

a closed interval because it contains both endpoints

and

(3.

The interval

containing neither endpoint, denoted by

(01.

(3),

consists of all x such that

< x <

(3;

this is called an open interval. And the intervals

[a..

(3)

and

(a. .

(31,

which contain just one endpoint, are defined similarly and called

half- open.

How many integers are contained in such intervals? The half-open inter-

vals are easier, so we start with them. In fact half-open intervals are almost

always nicer than open or closed intervals. For example, they’re additive-we

can combine the half-open intervals

[K.

(3)

and

[(3

. .

to form the half-open

interval [a. . y). This wouldn’t work with open intervals because the point

would be excluded, and it could cause problems with closed intervals because

would be included twice.

Back to our problem. The answer is easy if

and

are integers: Then

[(x..(3) containsthe (?-olintegers

01,

o~+l, . . . .

S-1,

assuming that 016 6.

Similarly (

0~.

(31

contains

integers in such a case. But our problem is

harder, because

and

are arbitrary reals. We can convert it to the easier

problem, though, since

when n is an integer, according to (3.7). The intervals on the right have

integer endpoints and contain the same number of integers as those on the left,

which have real endpoints. So the interval

[oL..

contains exactly [rjl

1~1

integers, and

(0~.

(31

contains

[(3]

La]. This is a case where we actually

want to introduce floor or ceiling brackets, instead of getting rid of them.

74 INTEGER FUNCTIONS

By the way, there’s a mnemonic for remembering which case uses floors

and which uses ceilings: Half-open intervals that include the left endpoint

but not the right (such as 0 <

< 1) are slightly more common than those

that include the right endpoint but not the left; and floors are slightly more

Just like we can

re-

common than ceilings. So by Murphy’s Law, the correct rule is the opposite

member

the date of

of what we’d expect -ceilings for

[OL

. . p) and floors for

(01.

01.

Columbus’s depar-

Similar analyses show that the closed interval

[o(.

. fi] contains exactly

t ure by singing,

“In

fourteen

hundred

Ll3J

[a] +1 integers and that the open interval (01..

contains [fi]

LX]- 1;

but we place the additional restriction

on the latter so that the formula

;o~u~~~-$;~;{~e

won’t ever embarrass us by claiming that an empty interval (a. . a) contains

deep

b,ue

sea

a total of -1 integers. To summarize, we’ve deduced the following facts:

interval integers contained restrictions

[a..

1B.l

Toil+1

[a..

I31

Ml - bl

(3.12)

(a.. Bl

LPJ

(a..B)

TPl

- 14 -1

Now here’s a problem we can’t refuse. The Concrete Math Club has a

casino (open only to purchasers of this book) in which there’s a roulette wheel

with one thousand slots, numbered 1 to 1000. If the number n that comes up

on a spin is divisible by the floor of its cube root, that is, if

then it’s a winner and the house pays us $5; otherwise it’s a loser and we

must pay $1. (The notation a\b, read “a divides

b,”

means that b is an exact

multiple of a; Chapter 4 investigates this relation carefully.) Can we expect

to make money if we play this game?

We can compute the average winnings-that is, the amount we’ll win

(or lose) per play-by first counting the number W of winners and the num-

ber L = 1000

W of losers. If each number comes up once during 1000 plays,

we win 5W dollars and lose L dollars, so the average winnings will be

[A poll of the class

this point showed

that 28 students

thought it was a

bad idea to play,

13 wanted to gam-

ble, and the rest

were too confused

5w-L

5w-(looo-w) 6W-

1000

to answer.)

1000 ;ooo

=1000 .

(So we hit them

with the Concrete

If there are 167 or more winners, we have the advantage; otherwise the ad-

Math

aub.1

vantage is with the house.

How can we count the number of winners among 1 through 1 OOO? It’s

not hard to spot a pattern. The numbers from 1 through

1 = 7 are all

winners because

[fi]

= 1 for each. Among the numbers

= 8 through

1 = 26, only the even numbers are winners. And among

= 27 through

1 = 63, only those divisible by 3 are. And so on.

3.2 FLOOR/CEILING APPLICATIONS 75

The whole setup can be analyzed systematically if we use the summa-

tion techniques of Chapter 2, taking advantage of

Iverson’s

convention about

logical statements evaluating to 0 or 1:

1000

n is a winner]

?I=1

[Lfij

\n] =

~[k=Lfi~][k\nl(l

6n610001

l<n61000

k,n

[k3$n<(k+1)3][n=km][l

6n<lOOO)

km,n

= 1

+~[k3<km<(k+l)3][l<k<10]

l+~[m~[k~..(k+1)~/k)][l~k<l0l

l+k’g

([k2+3k+3+l/kl-[k21)

l<k<lO

7+31

= 1+

(3k+4)

l+T.

9 = 172.

l<k<lO

nue.

Where did you say

this casino is?

This derivation merits careful study. Notice that line 6 uses our formula

(3.12) for the number of integers in a half-open interval. The only “difficult”

maneuver is the decision made between lines 3 and 4 to treat n = 1000 as a

special case. (The inequality

6 n < (k + 1

does not combine easily with

1 6 n < 1000 when k = 10.) In general, boundary conditions tend to be the

most critical part of x-manipulations.

The bottom line says that W = 172; hence our formula for average win-

nings per play reduces to (6.172

1000)/1000

dollars, which is 3.2 cents. We

can expect to be about $3.20 richer after making 100 bets of $1 each. (Of

course, the house may have made some numbers more equal than others.)

The casino problem we just solved is a dressed-up version of the more

mundane question, “How many integers n, where 1 6 n 6 1000, satisfy the re-

lation

LfiJ

\ n?” Mathematically the two questions are the same. But some-

times it’s a good idea to dress up a problem. We get to use more vocabulary

(like “winners” and “losers”), which helps us to understand what’s going on.

Let’s get general. Suppose we change 1000 to 1000000, or to an even

larger number, N . (We assume that the casino has connections and can get a

bigger wheel.) Now how many winners are there?

The same argument applies, but we need to deal more carefully with the

largest value of k, which we can call K for convenience:

76 INTEGER FUNCTIONS

(Previously K was 10.) The total number of winners for general N comes to

W =

(3k+4)

+x[K3<Km<N]

l<k<K

f(7+3K+l)(K~l)+~[mtlK2..N/K)]

$K2+sK-4+~[mE[K2..N/K]].

We know that the remaining sum is

LN/KJ

[K21

+ 1 = [N/K]

+ 1;

hence the formula

W =

LN/Kj+;K’+;K-3,

K =

[ml

(3.13)

gives the general answer for a wheel of size N.

The first two terms of this formula are approximately N2i3 + iN213 =

$N2j3, and the other terms are much smaller in comparison, when N is large.

In Chapter 9 we’ll learn how to derive expressions like

W = ;N2’3 +

O(N”3),

where O(N’j3) stands for a quantity that is no more than a constant times

N’13. Whatever the constant is, we know that it’s independent of N; so for

large N the contribution of the O-term to W will be quite small compared

with iN213. For example, the following table shows how close iN213 is to

p/3

W % error

1,000 150.0

172

12.791

10,000

696.2

746

6.670

100,000

3231.7

3343

3.331

1,000,000 15000.0

15247 1.620

1 o,ooo,ooo

69623.8

70158

0.761

100,000,000

323165.2 324322

0.357

1,000,000,000

1500000.0

1502496

0.166

It’s a pretty good approximation.

Approximate formulas are useful because they’re simpler than formu-

las with floors and ceilings. However, the exact truth is often important,

too, especially for the smaller values of N that tend to occur in practice.

For example, the casino owner may have falsely assumed that there are only

$N2j3 = 150 winners when N = 1000 (in which case there would be a

lO#

advantage for the house).

. . . without MS

of generality. .

“If x be

an in-

commensurable

number less than

unity, one of the

series of quantities

m/x, m/(1

-x),

where m is a whole

number, can be

found which shall

he between any

given consecutive

integers, and but

one such quantity

can be found.”

Rayleigh

[245]

Right, because

exact/y one of

the counts must

increase when n

increases

1 .

3.2 FLOOR/CEILING APPLICATIONS 77

Our last application in this section looks at so-called spectra. We define

the spectrum of a real number a to be an infinite multiset of integers,

Sped4

= 114,

12a1,

13a1, . . .I.

(A multiset is like a set but it can have repeated elements.) For example, the

spectrum of

l/2

starts out (0,

1,2,2,3,3,.

.}.

It’s easy to prove that no two spectra are equal-that a #

implies

Spec(a) # Spec((3). For, assuming without loss of generality that a <

(3,

there’s a positive integer m such that m(

a) 3

(In fact, any m 3

[l/(

a)] will do; but we needn’t show off our knowledge of floors and

ceilings all the time.) Hence ml3

ma 3 1, and LrnSl > [ma]. Thus

Spec((3) has fewer than m elements <

lrnaj,

while Spec(a) has at least m.

Spectra have many beautiful properties. For example, consider the two

multisets

Spec(&) =

{1,2,4,5,7,8,9,11,12,14,15,16,18,19,21,22,24

,...

Spec(2+fi)

{3,6,10,13,17,20,23,27,30,34,37,40,44,47,51,...

It’s easy to calculate Spec(

) with a pocket calculator, and the nth element

of Spec(2+

fi)

is just 2n more than the nth element of Spec(fi), by (3.6).

A closer look shows that these two spectra are also related in a much more

surprising way: It seems that any number missing from one is in the other,

but that no number is in both! And it’s true: The positive integers are the

disjoint union of Spec(

) and Spec(2+

We say that these spectra form

a partition of the positive integers.

To prove this assertion, we will count how many of the elements of

Spec(&!) are 6 n, and how many of the elements of

Spec(2+fi)

are 6 n. If

the total is n, for each n, these two spectra do indeed partition the integers.

Let a be positive. The number of elements in Spec(a) that are < n is

N(a,n) =

x[lkaJ

<n]

k>O

x[[kaj

<n+

k>O

ka<n+

k>O

x[O<k<(n+l)/a]

[;n+l)/a]

-1.

(3.14)

78 INTEGER FUNCTIONS

This derivation has two special points of interest. First, it uses the law

m<n

-e+

m<n+l,

integers m and n

(3.15)

to change ‘<’ to I<‘, so that the floor brackets can be removed by (3.7).

Also -and this is more subtle -it sums over the range k > 0 instead of k 3

because (n + 1 )/a might be less than 1 for certain n and a. If we had tried

to apply (3.12) to determine the number of integers in

. .

(n+

1)/a), rather

than the number of integers in (0.. (n+ 1)/a), we would have gotten the right

answer; but our derivation would have been faulty because the conditions of

applicability wouldn’t have been met.

Good, we have a formula for N (a, n). Now we can test whether or not

Spec(

) and Spec(Z+

) partition the positive integers, by testing whether

or not

N(fi,

n) + N(2 +

fi,

n) = n for all integers n > 0, using (3.14):

(3.2);

n+l

+2+JZ

(3.3).

Everything simplifies now because of the neat identity

i&=l;

our condition reduces to testing whether or not

{T}+(S)

= 1,

for all n > 0. And we win, because these are the fractional parts of

noninteger numbers that add up to the integer n + 1. A partition it is.

3.3

FLOOR/CEILING RECURRENCES

two

Floors and ceilings add an interesting new dimension to the study

of recurrence relations. Let’s look first at the recurrence

= 1;

k-+1

min(2K~,/2l,3K~,/3~),

for n 3 0.

(3.16)

Thus, for example,

is 1 + min(2Ko,3Ko) = 3; the sequence begins 1, 3, 3,

4, 7, 7, 7, 9, 9, 10, 13, . . . .

One of the authors of this book has modestly

decided to call these the Knuth numbers.

3.3 FLOOR/CEILING RECURRENCES 79

Exercise 25 asks for a proof or disproof that K, > n, for all n 3 0. The

first few K’s just listed do satisfy the inequality, so there’s a good chance that

it’s true in general. Let’s try an induction proof: The basis n = 0 comes

directly from the defining recurrence. For the induction step, we assume

that the inequality holds for all values up through some fixed nonnegative n,

and we try to show that K,+l > n + 1. From the recurrence we know that

n+l

minWl,pJ

,3Kln/31

The induction hypothesis tells us that

L,,/~J

3 2Ln/2J and

3Kln/3~

3 3 [n/31. However, 2[n/2J can be as small

as n

and 3 Ln/3J can be as small as n

2. The most we can conclude

from our induction hypothesis is that

Kn+l

> 1 + (n

2); this falls far short

of K,+l 3 n + 1.

We now have reason to worry about the truth of K, 3 n, so let’s try to

disprove it. If we can find an n such that either

2Kl,,zl

< n or 3Kl,,31 < n,

or in other words such that

we will have K,+j < n + 1. Can this be possible? We’d better not give the

answer away here, because that will spoil exercise 25.

Recurrence relations involving floors and/or ceilings arise often in com-

puter science, because algorithms based on the important technique of “divide

and conquer” often reduce a problem of size n to the solution of similar prob-

lems of integer sizes that are fractions of n. For example, one way to sort

n records, if n >

is to divide them into two approximately equal parts, one

of size [n/21 and the other of size Ln/2]. (Notice, incidentally, that

n = [n/21 + Ln/2J

;

(3.17)

this formula comes in handy rather often.) After each part has been sorted

separately (by the same method, applied recursively), we can merge the

records into their final order by doing at most n

1 further comparisons.

Therefore the total number of comparisons performed is at most f(n), where

f(1) = 0;

f(n)=f([n/21)+f([n/2J)+n-1,

for n > 1

(3.18)

A solution to this recurrence appears in exercise 34.

The

Josephus

problem of Chapter 1 has a similar recurrence, which can

be cast in the form

J(1) = 1;

J(n) = 2J( LnI2J)

(-1)” ,

for n > 1.

80 INTEGER FUNCTIONS

We’ve got more tools to work with than we had in Chapter 1, so let’s

consider the more authentic

Josephus

problem in which every third person is

eliminated, instead of every second. If we apply the methods that worked in

Chapter 1 to this more difficult problem, we wind up with a recurrence like

J3(n)

= [iJ3(Ljnl) + a,]

modn+

where ‘mod’ is a function that we will be studying shortly, and where we have

a,, = -2, +1 , or

-i

according as n mod 3 = 0,

or 2. But this recurrence

is too horrible to pursue.

There’s another approach to the

Josephus

problem that gives a much

better setup. Whenever a person is passed over, we can assign a new number.

Thus, 1 and 2 become n + 1 and n + 2, then 3 is executed; 4 and 5 become

3 and n

4, then 6 is executed; . . .

;

3kSl

and

3k+2

become

n+2k+

and n + 2k + 2, then 3k + 3 is executed; . . . then 3n is executed (or left to

survive). For example, when n = 10 the numbers are

12345678

9 10

11 12 13 14 15 16

19 20

23 24

26 27

The kth person eliminated ends up with number 3k. So we can figure out who

the survivor is if we can figure out the original number of person number 3n.

If N > n, person number N must have had a previous number, and we

can find it as follows: We have N = n + 2k + 1 or N = n + 2k + 2, hence

k = [(N

1)/2J

;

the previous number was 3k +

or 3k + 2, respectively.

That is, it was 3k + (N

2k) = k + N

n. Hence we can calculate the

survivor’s number

(n) as follows:

N := 3n;

while

N>n

do N:=

[“-r-‘]

+N-n;

J3(n)

:= N.

This is not a closed form for Jj(n); it’s not even a recurrence. But at least it

“Not too slow,

not

too

fast,”

tells us how to calculate the answer reasonably fast, if n is large. -L.

Amstrong

“Known” like, say,

harmonic numbers.

A. M.

Odlyzko and

H. S. Wilf have

shown that

D:’ =

[(

$)“Cj ,

where

1.622270503.

3.3 FLOOR/CEILING RECURRENCES 81

Fortunately there’s a way to simplify this algorithm if we use the variable

D = 3n + 1

N in place of N. (This change in notation corresponds to

assigning numbers from 3n down to

instead of from 1 up to 3n; it’s sort of

like a countdown.) Then the complicated assignment to N becomes

D :=

3n+l-

(3n+1-D)-n-1

+(3n+1-D)-n

and we can rewrite the algorithm as follows:

D := 1;

while D < 2n do D :=

[;Dl

;

Js(n) :=

3n+l

-D.

Aha!

This looks much nicer, because n enters the calculation in a very simple

way. In fact, we can show by the same reasoning that the survivor

(n) when

every qth person is eliminated can be calculated as follows:

D := 1;

while D

1)n do D :=

[*Dl

;

J,(n) :=

qn+l

-D.

(3.19)

In the case q = 2 that we know so well, this makes D grow to

2m+1

when

n==2”+1;

hence

Jz(n)=2(2m+1)+1

-2m+1

=21+1.

Good.

The recipe in (3.19) computes a sequence of integers that can be defined

by the following recurrence:

D(q)

D’4’

L,,(q)

n-1

for n >

(3.20)

These numbers don’t seem to relate to any familiar functions in a simple

way, except when q = 2; hence they probably don’t have a nice closed form.

But if we’re willing to accept the sequence

D$’

as “known,” then it’s easy to

describe the solution to the generalized

Josephus

problem: The survivor

(n)

qn+

-Dp’,

where k is as small as possible such that

D:’

> (q

1)n.

3.4 ‘MOD’: THE BINARY OPERATION

The quotient of n divided by m is Ln/m] , when m and n are positive

integers. It’s handy to have a simple notation also for the remainder of this

82 INTEGER FUNCTIONS

division, and we call it ‘n mod m’. The basic formula

n =

mLn/mJ

+ nmodm

quotient

remainder

tells us that we can express n mod m as n

mln/mJ

. We can generalize this

to negative integers, and in fact to arbitrary real numbers:

xmody = x

yLx/yJ,

for y # 0.

(3.21)

This defines ‘mod’ as a binary operation, just as addition and subtraction are

binary operations. Mathematicians have used mod this way informally for a

long time, taking various quantities mod 10, mod

277,

and so on, but only in

the last twenty years has it caught on formally. Old notion, new notation.

We can easily grasp the intuitive meaning of x mod y, when x and y

are positive real numbers, if we imagine a circle of circumference y whose

points have been assigned real numbers in the interval

. . y). If we travel a

distance x around the circle, starting at 0, we end up at x mod y. (And the

number of times we encounter 0 as we go is [x/y] .)

When x or y is negative, we need to look at the definition carefully in

order to see exactly what it means. Here are some integer-valued examples:

5mod3

5-3[5/3]

= 2;

5 mod -3

= 5

(-3)15/(-3)]

-1

;

-5 mod 3 = -5

3L-5/3]

= 1;

-5 mod -3

= -5

(-3)

l--5/(-3)]

= -2.

Why do they call it

‘mod’: The Binary

Operation? Stay

tuned to find out in

the next, exciting,

chapter!

Beware of computer

languages that

use

another definition.

The number after ‘mod’ is called the modulus; nobody has yet decided what

How about calling

to call the number before ‘mod’. In applications, the modulus is usually

positive, but the definition makes perfect sense when the modulus is negative.

:tz ~~~u~o~~

In both cases the value of x mod y is between 0 and the modulus:

0 < xmody < y, for y > 0;

0 2 xmody > y, for y < 0.

What about y = O? Definition (3.21) leaves this case undefined, in order to

avoid division by zero, but to be complete we can

define

xmod0

= x.

(3.22)

This convention preserves the property that x mod y always differs from x by

a multiple of y. (It might seem more natural to make the function continuous

at 0, by defining x mod 0 =

lim,,o

x mod y = 0. But we’ll see in Chapter 4

There was a time in

the 70s when ‘mod’

was the fashion.

Maybe the new

mumble function

should be called

‘punk’?

No-l

‘mumble’.

The remainder, eh?

3.4 ‘MOD’: THE BINARY OPERATION 83

that this would be much less useful. Continuity is not an important aspect

of the mod operation.)

We’ve already seen one special case of mod in disguise, when we wrote x

in terms of its integer and fractional parts, x =

1x1

+ {x}. The fractional part

can also be written x mod 1, because we have

x =

lxj

+ x mod 1 .

Notice that parentheses aren’t needed in this formula; we take mod to bind

more tightly than addition or subtraction.

The floor function has been used to define mod, and the ceiling function

hasn’t gotten equal time. We could perhaps use the ceiling to define a mod

analog like

xmumbley =

y[x/yl

-x;

in our circle analogy this represents the distance the traveler needs to continue,

after going a distance x, to get back to the starting point 0. But of course

we’d need a better name than ‘mumble’. If sufficient applications come along,

an appropriate name will probably suggest itself.

The distributive law is mod’s most important algebraic property: We

have

c(x mod y) = (cx) mod (cy)

(3.23)

for all real c, x, and y. (Those who like mod to bind less tightly than multi-

plication may remove the parentheses from the right side here, too.) It’s easy

to prove this law from definition (3.21), since

c(x mod y ) = c(x

y [x/y] ) = cx

cy [cx/cy] = cx mod cy ,

if cy # 0; and the zero-modulus cases are trivially true. Our four examples

using

and

illustrate this law twice, with c = -1. An identity like

(3.23) is reassuring, because it gives us reason to believe that ‘mod’ has not

been defined improperly.

In the remainder of this section, we’ll consider an application in which

‘mod’ turns out to be helpful although it doesn’t play a central role. The

problem arises frequently in a variety of situations: We want to partition

n things into m groups as equally as possible.

Suppose, for example, that we have n short lines of text that we’d like

to arrange in m columns. For aesthetic reasons, we want the columns to be

arranged in decreasing order of length (actually nonincreasing order); and the

lengths should be approximately the same-no two columns should differ by

84 INTEGER FUNCTIONS

more than one line’s worth of text. If 37 lines of text are being divided into

five columns, we would therefore prefer the arrangement on the right:

8 8

8 5 8 8 7 7 7

Furthermore we want to distribute the lines of text columnwise-first decid-

ing how many lines go into the first column and then moving on to the second,

the third, and so on-because that’s the way people read. Distributing row

by row would give us the correct number of lines in each column, but the

ordering would be wrong. (We would get something like the arrangement on

the right, but column 1 would contain lines 1, 6, 11, . . . , 36, instead of lines

1, 2, 3, . .

)

as desired.)

A row-by-row distribution strategy can’t be used, but it does tell us how

many lines to put in each column. If n is not a multiple of m, the

row-

by-row procedure makes it clear that the long columns should each contain

[n/ml lines, and the short columns should each contain

Ln/mJ.

There will

be exactly n mod m long columns (and, as it turns out, there will be exactly

n mumble m short ones).

Let’s generalize the terminology and talk about ‘things’ and ‘groups’

instead of ‘lines’ and ‘columns’.

We have just decided that the first group

should contain [n/ml things; therefore the following sequential distribution

scheme ought to work: To distribute n things into m groups, when m > 0,

put [n/ml things into one group, then use the same procedure recursively to

put the remaining n’ = n- [n/ml things into m’ = m- 1 additional groups.

For example, if n = 314 and m = 6, the distribution goes like this:

remaining things remaining groups [things/groups]

314 6 53

261

208 4 52

156

104 2 52

It works. We get groups of approximately the same size, even though the

divisor keeps changing.

Why does it work? In general we can suppose that n = qm +

where

q =

Ln/mJ

and r = n mod m. The process is simple if r = 0: We put

[n/ml = q things into the first group and replace n by n’ = n

q, leaving

3.4 ‘MOD’: THE BINARY OPERATION 85

n’ = qm’ things to put into the remaining m’ = m

1 groups. And if

> 0, we put

[n/ml

= q + 1 things into the first group and replace n

by n’ = n

leaving n’ = qm’ +

1 things for subsequent groups.

The new remainder is r’ =

but q stays the same. It follows that there

will be

groups with q + 1 things, followed by m

groups with q things.

How many things are in the kth group? We’d like a formula that gives

[n/ml

when k < n mod m, and Ln/m] otherwise. It’s not hard to verify

that

has the desired properties, because this reduces to q + [(r

k + 1 )/ml if we

write n = qm +

as in the preceding paragraph; here q =

[n/m].

We have

[(r-k+

1)/m]

[k<r],

if 1 6 k 6 m and 0 6

< m. Therefore we can

write an identity that expresses the partition of n into m as-equal-as-possible

parts in nonincreasing order:

This identity is valid for all positive integers m, and for all integers n (whether

positive, negative, or zero). We have already encountered the case m = 2 in

(3.17), although we wrote it in a slightly different form, n = [n/21 +

[n/2].

If we had wanted the parts to be in nondecreasing order, with the small

groups coming before the larger ones, we could have proceeded in the same

way but with

[n/mJ

things in the first group. Then we would have derived

the corresponding identity

(3.25)

It’s possible to convert between (3.25) and (3.24) by using either (3.4) or the

identity of exercise 12.

Some c/aim that it’s

Now if we replace n in (3.25) by Lrnx] , and apply rule (3.11) to remove

too dangerous to

replace anything by

floors inside of floors, we get an identity that holds for all real x:

an mx.

LmxJ =

1x1

..+

lx+&J]

+ -!- +

(3.26)

This is rather amazing, because the floor function is an integer approximation

of a real value, but the single approximation on the left equals the sum of a

bunch of them on the right. If we assume that

1x1

is roughly x

on the

average, the left-hand side is roughly mx

5, while the right-hand side comes

toroughly

(x--)+(x-it-l-)+...

+(x-i+%)

=mx-it;

thesumof

all these rough approximations turns out to be exact!

86 INTEGER FUNCTIONS

3.5 FLOOR/CEILING SUMS

Equation (3.26) demonstrates that it’s possible to get a closed form

for at least one kind of sum that involves

J. Are there others? Yes. The

trick that usually works in such cases is to get rid of the floor or ceiling by

introducing a new variable.

For example, let’s see if it’s possible to do the sum

in closed form. One idea is to introduce the variable m = L&J; we can do

this “mechanically” by proceeding as we did in the roulette problem:

l&J

m[k<nl[m=lfil]

O<k<n k,m>O

m[k<nl[m<fi<m+l

k.m>O

m[k<nl[m2<k<(m+1

)‘I

m[m2<k<(m+1)2<n]

m[mLSk<n<(m+1)2]

k,m>O

Once again the boundary conditions are a bit delicate. Let’s assume first that

n =

is a perfect square. Then the second sum is zero, and the first can be

evaluated by our usual routine:

k,m>O

tm((m+l)‘-m2)[m+16al

ll@O

~m(2m+l)[m<al

Ill20

(2mZ+3ml)[m<

ll@O

x,”

(2mL +

3ml)

$a(a-l)(a-2)+$a(a-1)

;(4a+l)a(a-1).

Falling powers

make the sum come

tumbling down.

Warning: This stuff

is fairly advanced.

Better skim the

next two pages on

first reading; they

aren't crucial.

-Friendly TA

Start

Skimming

3.5 FLOOR/CEILING SUMS 87

In the general case we can let a =

Lfij;

then we merely need to add

the terms for

< k < n, which are all equal to a, so they sum to (n

a2)a.

This gives the desired closed form,

lJi;J

na-ia3-ia2-ia,

[J;;J.

(3.27)

O<k<n

Another approach to such sums is to replace an expression of the form

1x1

,‘Yj

[l $ j 6 xl; this is legal whenever x 3 0. Here’s how that method

works in the sum of [square rodts], if we assume for convenience that n = a2:

l&j

~[1<j~&l[06k<a21

O<k<n

‘5

~[j2<k<a2]

l<j<a

(a’-j2)

fa(a+

:)(a+

1).

l<j<a

Now here’s another example where a change of variable leads to a trans-

formed sum. A remarkable theorem was discovered independently by three

mathematicians- Bohl

[28],

Sierpiliski

[265],

and Weyl

[300]

-at about the

same time in 1909: If

is irrational then the fractional parts {na} are very uni-

formly distributed between 0 and

as n

00. One way to state this is that

)im;

f({ka}) = 1; f(x) dx

O<k<n

(3.28)

for all irrational

and all functions f that are continuous almost everywhere.

For example, the average value of

{TUX}

can be found by setting f(x) = x; we

get

(That’s exactly what we might expect; but it’s nice to know that it is

really, provably true, no matter how irrational

is.)

The theorem of Bohl, Sierpifiski, and Weyl is proved by approximating

f(x) above and below by “step functions,’ which are linear combinations of

the simple functions

f"(X) =

[06x<vl

when 0 < v 6 1. Our purpose here is not to prove the theorem; that’s a job

for calculus books. But let’s try to figure out the basic reason why it holds,

by seeing how well it works in the special case f(x) = f,,(x). In other words,

let’s try to see how close the sum

O<k<n

gets to the “ideal” value nv, when n is large and

is irrational.

88 INTEGER FUNCTIONS

For this purpose we define the discrepancy D(ol,n) to be the maximum

absolute value, over all 0 6 v <

of the sum

s(a,n,v) =

([{ka}<v]

-v).

O<k<n

(3.29)

Our goal is to show that D(

LX,

n) is “not too large” when compared with n,

by showing that

Is(a,

n,v)l is always reasonably small.

First we can rewrite s(a, n,v) in simpler form, then introduce a new

index variable j:

([{ka}<v] -v) =

([ka]

-[klx-VI-v)

O<k<n O<k<n

-nv+

ELka--vvjjka]

O<k<n j

= -nv+

[jaP’<k<(j+v)a-‘1.

O<j<rna]

kin

If we’re lucky, we can do the sum on k. But we ought to introduce some

new variables, so that the formula won’t be such a mess. Without loss of

generality, we can assume that 0 < a < 1; let us write

Right, name and

conquer.

a = ~ap’J , a-’ =

a+a’;

The change of

vari-

able from k to j is

b = [va-‘l ,

va-’ = b -v’.

Thus a’ = {a--‘} is the fractional part of a-‘, and v’ is the mumble-fractional

part of

va-‘.

Once again the boundary conditions are our only source of grief. For

now, let’s forget the restriction ‘k < n’ and evaluate the sum on k without it:

[kc

[ja-’

..(j+v)a-‘)I

+ v)(a + a’)]

[j(a + a’)]

= b+

[ja’-v’l

[ja’l.

OK, that’s pretty simple; we plug it in and plug away:

s(a,n,v) = -nv+

1nalb-t

([ja’-v’l

[ja’l)

-S,

(3.30)

O<j<[nal

the main point.

Friendly TA

where S is a correction for the cases with k 3 n that we have failed to exclude.

The quantity ja’ will never be an integer, since a (hence a’) is irrational; and

ja’

-v’

will be an integer for at most one value of j. So we can change the

3.5 FLOOR/CEILING SUMS 89

ceiling terms to floors:

(The formula

[O or 1

stands

for something that’s

either 0 or 1

;

needn’t commit

ourselves, because

the details don’t

really matter.)

E2ming

s(oI,n,v) =

-nv+[noilb-

(Lja’J-LjoL’-v’J)-S+[Oor

11.

O<j<

[nal

Interesting. Instead of a closed form, we’re getting a sum that looks rather

s(oI,

n, v) but with different parameters:

LX’

instead of

K, [no;]

instead

of n, and v’ instead of v. So we’ll have a recurrence for

01,

n,v),

which

(hopefully) will lead to a recurrence for the discrepancy D

(01,

n). This means

we want to get

s(oI’,

[noil,v’) =

(lja’j

ljcx-v’j

-v’)

O<ji[nal

into the act:

s(oL,n,v) = -nv+

[nalb-

[nOiJv’-s(a’,[nOil,v’)-S+[Oor

11.

Recalling that b

-v’

VK’

, we see that everything will simplify beautifully

if we replace [na] (b

v’) by nol(b

-v’)

= nv:

s(ol,n,v) = -S(K), [nO(l,v’) -S +

or 11.

Here

is a positive error of at most

VOL-‘.

Exercise 18 proves that S is,

likewise, between 0 and 01-l. We can also remove the term for j = [n&l

1 =

[n.K]

from the sum, since it contributes either v’ or v’

1. Hence, if we take

the maximum of absolute values over all v, we get

D(ol,n) < D(oI’, [KnJ) +

0~~’

$2.

(3.31)

The methods we’ll learn in succeeding chapters will allow us to conclude

from this recurrence that D(ol,n) is always much smaller than n, when n is

sufficiently large. Hence the theorem (3.28) is not only true, it can also be

strengthened: Convergence to the limit is very fast.

Whew; that was quite an exercise in manipulation of sums, floors, and

ceilings. Readers who are not accustomed to “proving that errors are small”

might find it hard to believe that anybody would have the courage to keep

going, when faced with such weird-looking sums. But actually, a second look

shows that there’s a simple motivating thread running through the whole

calculation. The main idea is that a certain sum

s(01,

n,v) of n terms can be

reduced to a similar sum of at most

oLn

terms. Everything else cancels out

except for a small residual left over from terms near the boundaries.

Let’s take a deep breath now and do one more sum, which is not trivial

but has the great advantage (compared with what we’ve just been doing) that

90 INTEGER FUNCTIONS

it comes out in closed form so that we can easily check the answer. Our goal

now will be to generalize the sum in (3.26) by finding an expression for

Is this a harder

sur’n

of floors, or a sum

of harder floors?

integer m > 0, integer n.

Finding a closed form for this sum is tougher than what we’ve done so far

(except perhaps for the discrepancy problem we just looked at). But it’s

Be forewarned: This

instructive, so we’ll hack away at it for the rest of this chapter.

is the beginning of

As usual, especially with tough problems, we start by looking at small

a pattern, in that

cases. The special case n = 1 is (3.26), with x replaced by x/m:

the

last

part

the

chapter consists

of ihe solution of

And as in Chapter 1, we find it useful

downwards to the case n = 0:

LXJ

some long, difficult

problem, with little

more motivation

to get more data by generalizing

than curiosity.

-Students

Touch&

But c’mon,

gang, do you always

need to be to/d

about applications

before you can get

interested in some-

thing? This sum

arises, for example,

in the study of

random number

generation and

testing. But math-

ematicians looked

at it long before

computers came

along, because they

found it natural to

ask if there’s a way

to sum arithmetic

progressions that

have been “floored.”

Our problem has two parameters, m and n; let’s look at some small cases

for m. When m = 1 there’s just a single term in the sum and its value is

1x1.

When m = 2 the sum is 1x/2] + [(x +

n)/2J.

We can remove the interaction

between x and n by removing n from inside the floor function, but to do that

we must consider even and odd n separately. If n is even, n/2 is an integer,

so we can remove it from the floor:

If n is odd, (n

1)/2 is an integer so we get

The last step follows from (3.26) with m = 2.

These formulas for even and odd n slightly resemble those for n = 0 and

but no clear pattern has emerged yet; so we had better continue exploring

some more small cases. For m = 3 the sum is

-Your instructor

and we consider three cases for n: Either it’s a multiple of 3, or it’s 1 more

than a multiple, or it’s 2 more. That is, n mod 3 = 0,

or 2. If n mod 3 = 0

3.5 FLOOR/CEILING SUMS 91

then n/3 and 2n/3 are integers, so the sum is

“inventive genius

requires pleasurable

mental activity as

a condition for its

vigorous exercise.

‘Necessity is the

mother of invention’

is a silly proverb.

‘Necessity is the

mother of futile

dodges’is

much

nearer to the truth.

The basis of the

growth of modern

invention is science,

and science is al-

most wholly the

outgrowth of plea-

surable intellectual

curiosity.”

-A. N.

White-

head [303]

If n mod 3 = 1 then (n

1)/3 and (2n

2)/3 are integers, so we have

Again this last step follows from (3.26), this time with m = 3. And finally, if

n mod 3 = 2 then

The left hemispheres of our brains have finished the case m = 3, but the

right hemispheres still can’t recognize the pattern, so we proceed to m = 4:

At least we know enough by now to consider cases based on n mod m. If

n mod 4 = 0 then

Andifnmod4=1,

The case n mod 4 = 3 turns out to give the same answer. Finally, in the case

n mod 4 = 2 we get something a bit different, and this turns out to be an

important clue to the behavior in general:

This last step simplifies something of the form [y/2] + [(y + 1)/2J, which

again is a special case of (3.26).

92 INTEGER FUNCTIONS

To summarize, here’s the value of our sum for small m:

nmodm=O

nmodm=l

nmodm=2

nmodm=3

3[:]+n

1x1

+ n

LxJ

+ n

It looks as if we’re getting something of the form

where a, b, and c somehow depend on m and n. Even the myopic among

us can see that b is probably (m

1)/2.

It’s harder to discern an expression

for a; but the case n mod 4 = 2 gives us a hint that a is probably gcd(m, n),

the greatest common divisor of m and n. This makes sense because gcd(m, n)

is the factor we remove from m and n when reducing the fraction n/m to

lowest terms, and our sum involves the fraction n/m. (We’ll look carefully

at gcd operations in Chapter 4.) The value of c seems more mysterious, but

perhaps it will drop out of our proofs for a and b.

In computing the sum for small m, we’ve effectively rewritten each term

of the sum as

because (kn

kn mod m)/m is an integer that can be removed from inside

the floor brackets. Thus the original sum can be expanded into the following

tableau:

Omodm

m m m

nmodm

2n 2n mod m

x+(m-1)nmodm

(m-lb

(m-l)nmodm

m m m

3.5 FLOOR/CEILING SUMS 93

When we experimented with small values of m, these three columns led re-

spectively to a[x/aJ, bn, and c.

In particular, we can see how b arises. The second column is an arithmetic

progression, whose sum we know-it’s the average of the first and last terms,

times the number of terms:

;o+

(

(m- 1)n

.m =

(m-lb

So our guess that b = (m

1)/2 has been verified.

The first and third columns seem tougher; to determine a and c we must

take a closer look at the sequence ofnumbers

Omodm, nmodm,

2nmodm,

. . . . (m-1)nmodm.

Lemmanow,

dilemma

later.

Suppose, for example, that m = 12 and n = 5. If we think of the

sequence as times on a clock, the numbers are 0 o’clock (we take 12 o’clock

to be 0 o’clock), then 5 o’clock, 10 o’clock, 3 o’clock (= 15 o’clock), 8 o’clock,

and so on. It turns out that we hit every hour exactly once.

Now suppose m = 12 and n = 8. The numbers are 0 o’clock, 8 o’clock,

4 o’clock (= 16 o’clock), but then 0, 8, and 4 repeat. Since both 8 and 12 are

multiples of 4, and since the numbers start at 0 (also a multiple of

4),

there’s

no way to break out of this pattern-they must all be multiples of 4.

In these two cases we have gcd( 12,5) = 1 and gcd( 12,8) = 4. The general

rule, which we will prove next chapter, states that if d = gcd(m,n) then we

get the numbers 0, d,

2d,

. . . ,

d in some order, followed by d

1 more

copies of the same sequence. For example, with m = 12 and n = 8 the pattern

0, 8, 4 occurs four times.

The first column of our sum now makes complete sense. It contains

d copies of the terms [x/m],

1(x

+ d)/mJ, . . . ,

1(x

+ m

d)/m], in some

order, so its sum is

This last step is yet another application of (3.26). Our guess for a has been

verified:

a = d = gcd(m, n)

94 INTEGER FUNCTIONS

Also, as we guessed, we can now compute c, because the third column

has become easy to fathom. It contains d copies of the arithmetic progression

O/m, d/m, 2d/m, . , (m

d)/m, so its sum is

d(;(()+!$).$

the third column is actually subtracted, not added, so we have

d-m

c =

End of mystery, end of quest. The desired closed form is

where d = gcd(m, n). As a check, we can make sure this works in the special

cases n = 0 and n = 1 that we knew before: When n = 0 we get d =

gcd(m,O)

= m; the last two terms of the formula are zero so the formula

properly gives mLx/ml. And for n = 1 we get d = gcd(m, 1) = 1; the last

two terms cancel nicely, and the sum is just

1x1.

By manipulating the closed form a bit, we can actually make it symmetric

in m and n:

[T/

=d[???+~n+!$-?!

O$k<m

(m-l)(n-1) + m-l + d-m

2 2 2

(3.32)

This is astonishing, because there’s no reason to suspect that such a sum

Yup, I’m floored.

should be symmetrical. We have proved a “reciprocity law,’

For example, if m = 41 and n = 127, the left sum has 41 terms and the right

has 127; but they still come out equal, for all real x.

You know you’re

in college when the

book doesn’t tell

you how

to pro-

nounce

‘Dirichlet’.

3 EXERCISES 95

Exercises

Warmups

When we analyzed the

Josephus

problem in Chapter 1, we represented

an arbitrary positive integer n in the form n =

where 0 <

< 2”.

Give explicit formulas for

and m as functions of n, using floor and/or

ceiling brackets.

What is a formula for the nearest integer to a given real number x? In case

of ties, when x is exactly halfway between two integers, give an expression

that rounds (a) up-that is, to [xl; (b) down-that is, to

Lx].

Evaluate

\m&]n/a] , h

en m and n are positive integers and a is an

irrational number greater than n.

The text describes problems at levels 1 through 5. What is a level 0

problem? (This, by the way, is not a level 0 problem.)

Find a necessary and sufficient condition that

LnxJ

= n[xJ , when n is a

positive integer. (Your condition should involve {x}.)

Can something interesting be said about

Lf(x)J

when f(x) is a continuous,

monotonically decreasing function that takes integer values only when

x is an integer?

‘7

Solve the recurrence

X, = n, for 0 6 n < m;

x, =

x,-,+1,

for n 3 m.

Prove the Dirichlet box principle: If n objects are put into m boxes,

some box must contain 3 [n/ml objects, and some box must contain

lnhl.

Egyptian mathematicians in 1800 B.C. represented rational numbers be-

tween 0 and 1 as sums of unit fractions 1

/xl

. . . + 1

/xk,

where the x’s

were distinct positive integers. For example, they wrote

instead

Prove that it is always possible to do this in a systematic way: If

O<m/n<l,then

m 1

-=-

representation

n 4

4’

(This is Fibonacci’s algorithm, due to Leonardo Fibonacci, A.D. 1202.)

96 INTEGER FUNCTIONS

Basics

10 Show that the expression

is always either

1x1

or [xl. In what circumstances does each case arise?

11 Give details of the proof alluded to in the text, that the open interval

(a..

(3)

contains exactly

[(31

[a]

1 integers when a <

l3.

Why does

the case a =

have to be excluded in order to make the proof correct?

12 Prove that

n+m-1

’

for all integers n and all positive integers m. [This identity gives us

another way to convert ceilings to floors and vice versa, instead of using

the reflective law

(3.4).]

Let a and

be positive real numbers. Prove that Spec(a) and

Spec(

partition the positive integers if and only if a and

are irrational and

l/a+l/P

=l.

14 Prove or disprove:

(xmodny)mody = xmody, integer n.

15 Is there an identity analogous to (3.26) that uses ceilings instead of floors?

Prove that n mod 2 = (1

(-1)“) /2. Find and prove a similar expression

for

mod 3 in the form a

bw” +

CW~“,

where w is the complex number

(-1

+i&)/2.

Hint:

cu3

= 1 and 1

+w+w’=O.

17 Evaluate the sum &Gk<m lx + k/mJ in the case x 3 0 by substituting

< j < x + k/m] for lx + k/m] and summing first on k. Does your

answer agree with

(3.26)?

18 Prove that the boundary-value error term S in (3.30) is at most a-Iv.

Hint: Show that small values of j are not involved.

Homework exercises

Find a necessary and sufficient condition on the real number b > 1 such

that

for all real x 3 1.

3 EXERCISES 97

20 Find the sum of all multiples of x in the closed interval

[(x..

fi],

when

x > 0.

21 How many of the numbers

2",

for 0 6 m < M, have leading digit 1 in

decimal notation?

22 Evaluate the sums S, =

[n/2k +

and

tk3,

[n/2k

+ i]

Show that the nth element of the sequence

1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...

[fi

51.

(The sequence contains exactly m occurrences of m.)

24 Exercise 13 establishes an interesting relation between the two multisets

Spec(oL)

and Spec(oc/(ol- l)), when

is any irrational number > 1,

because 1

/OL

+ (

)/OL

= 1. Find (and provej an interesting relation

between the two multisets Spec(a) and Spec(oL/(a+ l)), when

is any

positive real number.

25 Prove or disprove that the Knuth numbers, defined by

(3.16),

satisfy

3 n for all nonnegative n.

26 Show that the auxiliary

Josephus

numbers (3.20) satisfy

for n 3 0.

27 Prove that infinitely many of the numbers

DF’

defined by (3.20) are

even, and that infinitely many are odd.

28 Solve the recurrence

= 1;

an-l

lJan-l.l,

for n > 0.

29 Show that, in addition to

(3.31),

we have

D(oL,n)

D(oI’,

1an.J)

0~~’

-2.

30 Show that the recurrence

= m,

x:-,-2,

for n > 0,

has the solution X, =

[01~“1,

if m is an integer greater than

a +

0~~’

= m and

For example, if m = 3 the solution is

where

[@2n+’

)

l+Js

4=-y-,

a =

a2.

98 INTEGER FUNCTIONS

31 Prove or disprove:

1x1

\yJ

12x1

[ZyIJ

Let

(Ix((

= min(x

1x1,

[xl

-x)

denote the distance from x to the nearest

integer. What is the value of

2kllx/2kJJ2

(Note that this sum can be doubly infinite. For example, when x =

l/3

the terms are nonaero as k

-oo and also as k

+oo.)

Exam problems

33 A circle, 2n

1 units in diameter, has been drawn symmetrically on a

2n x 2n chessboard, illustrated here for n = 3:

How many cells of the board contain a segment of the circle?

b Find a function f(n, k) such that exactly

xc::

f(n, k) cells of the

board lie entirely within the circle.

34 Let f(n) =

Et=,

[lgkl.

Find a closed form for f(n) , when n 3 1.

Provethatf(n)=n-l+f([n/2~)+f(~n/Z])foralln~l.

Simplify the formula \(n + 1

)‘n!

mod n.

Simplify it,

but

36 Assuming that n is a nonnegative integer, find a closed form for the sum

don’t change the

value,

l<k<Z2”

2lk“J4lkkkJ

37 Prove the identity

(Lm-Jkj

1:~)

[:J

jmi+mOdn;lim)

mOdn12J

O$k<m

for all positive integers m and

38 Let

x1,

.,.,

be real numbers such that the identity

holds for all positive integers m. Prove something interesting about

Xl,

.‘.) x,.

3 EXERCISES 99

39 Prove that the double sum &k~‘og,x

&j<b[(~

+ jbk)/bk+‘] equals

(b- l)(Llog’,xl + 1) +

[xl

or every real number x 3 1 and every

integer b > 1.

40 The spiral function o(n), indicated in the diagram below, maps a non-

negative integer n onto an ordered pair of integers (x(n), y (n)). For

example, it maps n = 9 onto the ordered pair

(1,2).

a Prove that if m = [J;;I,

x(n) =

(-l)“((n-m(m+l)).[[ZfiJ

iseven] +

[irnl),

and find a similar formula for y(n). Hint: Classify the spiral into

segments

Wk,

Sk,

Ek,

according as

[2fij

= 4k

2, 4k

4k,

4k+

b Prove that, conversely, we can determine n from o(n) by a formula

of the form

WI2

(2k+x(n) +y(n))

m=(lx(n)l,lv(n)l).

Give a rule for when the sign is + and when the sign is

Bonus problems

41 Let f and g be increasing functions such that the sets {f

(1))

f (2), . . .

}

and

{g (1) , g

(2))

. .

}

partition the positive integers. Suppose that f and g are

related by the condition g(n) = f(f(n)) + 1 for all n > 0. Prove that

f(n) =

[n@J

and g(n) =

ln@‘J,

where

= (1 +

&)/2.

42 Do there exist real numbers a,

(3,

and y such that Spec(a),

Spec(

(3),

and

Spec(y)

together partition the set of positive integers?

100 INTEGER FUNCTIONS

43 Find an interesting interpretation of the Knuth numbers, by unfolding

the recurrence (3.16).

44 Show that there are integers

aiq’

and

diq)

such that

ac4)

D;!,

diq)

D(q)

d(q)

n n

q-l

’

for n > 0,

when

DIP’

is the solution to (3.20). Use this fact to obtain another form

of the solution to the generalized

Josephus

problem:

(n) = 1 +

d(‘)

+ q(n

aCq))

’

for ap’ 6 n <

ctp>“,‘,

45 Extend the trick of exercise 30 to find a closed-form solution to

= m,

= 2Yip,

)

for n > 0,

if m is a positive integer.

46 Prove that if n =

fi’

fi’-‘)mi

, where m and

are nonnegative

integers, then

Ld-1

l(&!“’

fi’)rnl

. Use this remarkable

property to find a closed form solution to the recurrence

= a, integer a > 0;

[-\/2LndL-l

+l)],

for

Hint:

[&Gi$ZXJ]

[Jz(n

+ t)J.

47 The function f(x) is said to be replicative if it satisfies

f(mx)

= f(x) +f(x+

+...+f(x+

for every positive integer m. Find necessary and sufficient conditions on

the real number c for the following functions to be replicative:

f(x) = x + c.

b f(x) = [x + c is an integer].

c f(x) =max([xJ,c).

d f(x) = x + c

1x1

is not an integer].

48 Find a necessary and sufficient condition on the real numbers 0 6

and

3 0 such that we can determine

and

from the infinite multiset

of values

{

Inal

(

3 EXERCISES 101

Research problems

Find a necessary and sufficient condition on the nonnegative real numbers

a and

such that we can determine a and

from the infinite multiset

of values

bet x be a real number 3

(1 +

&).

The solution to the recurrence

Zo(x)

Z,(x) =

Z,&x)'-1

for n > 0,

can be written Z,(x) =

[f(x)2”1,

if x is an integer, where

f(x) = $nmZn(x)1'2n ,

because Z,(x)

1 < f

(x)2”

< Z,(x). What interesting properties does

this function

f(x)

have?

Given nonnegative real numbers

and

(3,

let

Sw(a;P)

{la+PJ,l2a+P1,13a+P1,...}

be a multiset that generalizes Spec(a) =

Spec(a;

0). Prove or disprove:

If the m 3 3 multisets

Spec(a1;

PI),

Spec(a2;

/32),

. . . , Spec(a,;

&,,)

partition the positive integers, and if the parameters

’

. . < a,,,

are rational, then

2m-1

2k-1

’

for

Fibonacci’s algorithm (exercise 9) is “greedy” in the sense that it chooses

the least conceivable q at every step. A more complicated algorithm is

known by which every fraction m/n with n odd can be represented as a

sum of distinct unit fractions

/qj

+ .

/qk

with odd denominators.

Does the greedy algorithm for such a representation always terminate?

Number Theory

INTEGERS ARE CENTRAL to the discrete mathematics we are emphasiz-

ing in this book. Therefore we want to explore the theory of numbers, an

important branch of mathematics concerned with the properties of integers.

We tested the number theory waters in the previous chapter, by intro-

ducing binary operations called ‘mod’ and ‘gcd’. Now let’s plunge in and

really immerse ourselves in the subject.

4.1 DIVISIBILITY

We say that m divides n (or n is divisible by m) if m > 0 and the

ratio

n/m

is an integer. This property underlies all of number theory, so it’s

convenient to have a special notation for it. We therefore write

m\n

m > 0 and n = mk for some integer k.

(4.1)

(The notation ‘mln’ is actually much more common than ‘m\n’ in current

mathematics literature. But vertical lines are overused-for absolute val-

ues, set delimiters, conditional probabilities, etc. -and backward slashes are

underused. Moreover, ‘m\n’ gives an impression that m is the denominator of

an implied ratio. So we shall boldly let our divisibility symbol lean leftward.)

If m does not divide n we write

‘m!qn’.

There’s a similar relation, “n is a multiple of

m,”

which means almost

the same thing except that m doesn’t have to be positive. In this case we

simply mean that n = mk for some integer k. Thus, for example, there’s only

one multiple of 0 (namely 0), but nothing is divisible by 0. Every integer is

a multiple of -1, but no integer is divisible by -1 (strictly speaking). These

definitions apply when m and n are any real numbers; for example,

271

divisible by

7~.

But we’ll almost always be using them when m and n are

integers. After all, this is number theory.

102

other words, be

prepared to drown.

‘I

no integer is

dksible by -1

(strictly speaking).”

-Graham,

Knuth,

and Patashnik [131]

In Britain we call

this

‘hcf’

(highest

common factor).

Not to be confused

with the greatest

common multiple.

(Remember that

m’ or n’ can be

negative.)

4.1 DIVISIBILITY

103

The greatest common divisor of two integers m and n is the largest

integer that divides them both:

gcd(m,n) = max{ k 1 k\m and

k\n}.

(4.2)

For example, gcd( 12,lS) = 6. This is a familiar notion, because it’s the

common factor that fourth graders learn to take out of a fraction m/n when

reducing it to lowest terms: 12/18 = (12/6)/( 1 S/6) = 2/3. Notice that if

n > 0 we have gcd(0, n) = n, because any positive number divides 0, and

because n is the largest divisor of itself. The value of gcd(0,O) is undefined.

Another familiar notion is the

least

common multiple,

lcm(m,n) = min{k 1

k>O,

m\k and

n\k};

(4.3)

this is undefined if m < 0 or n 6 0. Students of arithmetic recognize this

as the least common denominator, which is used when adding fractions with

denominators m and n. For example, lcm( 12,lS) = 36, and fourth graders

know that

+ & =

The lcm is somewhat analogous to the

gcd, but we don’t give it equal time because the gcd has nicer properties.

One of the nicest properties of the gcd is that it is easy to compute, using

a 2300-year-old method called Euclid’s algorithm. To calculate gcd(m,n),

for given values 0 < m < n, Euclid’s algorithm uses the recurrence

gcd(O,n) = n;

gcd(m,n) = gcd(n mod m, m) , for m > 0.

(4.4)

Thus, for example, gcd( 12,lS) = gcd(6,12) = gcd(0,6) = 6. The stated

recurrence is valid, because any common divisor of m and n must also be a

common divisor of both m and the number n mod m, which is n

[n/m]

There doesn’t seem to be any recurrence for lcm(m,n) that’s anywhere near

as simple as this. (See exercise 2.)

Euclid’s algorithm also gives us more: We can extend it so that it will

compute integers m’ and n’ satisfying

m’m + n’n = gcd(m, n) .

(4.5)

Here’s how. If m = 0, we simply take m’ = 0 and n’ = 1. Otherwise we

let

= n mod m and apply the method recursively with

and m in place of

m and n, computing

and

ii%

such that

Fr +

?%rn

= gcd(r, m) .

Since

= n

[n/m]m

and gcd(r, m) = gcd(m,n), this equation tells us that

Y(n-

ln/mJm)

+mm

gcd(m,n).

104 NUMBER THEORY

The left side can be rewritten to show its dependency on m and n:

(iTi

[n/mj

m + Fn = gcd(m, n)

;

hence m’ =

[n/mJF and n’ = f are the integers we need in (4.5). For

example, in our favorite case m = 12, n =

18,

this method gives 6 = 0.0+1.6 =

1.6+0+12=(-1).12+1.18.

But why is (4.5) such a neat result? The main reason is that there’s a

sense in which the numbers m’ and n’ actually prove that Euclid’s algorithm

has produced the correct answer in any particular case. Let’s suppose that

our computer has told us after a lengthy calculation that gcd(m, n) = d and

that m’m + n’n =

but we’re skeptical and think that there’s really a

greater common divisor, which the machine has somehow overlooked. This

cannot be, however, because any common divisor of m and n has to divide

m’m + n’n; so it has to divide d; so it has to be 6 d. Furthermore we can

easily check that d does divide both m and n. (Algorithms that output their

own proofs of correctness are called

self-cetiifiing.)

We’ll be using (4.5) a lot in the rest of this chapter. One of its important

consequences is the following mini-theorem:

k\m and k\n

&Cm,

(4.6)

(Proof: If k divides both m and n, it divides m’m + n’n, so it divides

gcd( m, n) . Conversely, if k divides gcd( m,

n),

it divides a divisor of m and a

divisor of n, so it divides both m and n.) We always knew that any common

divisor of m and n must be less than or equal to their gcd; that’s the

definition of greatest common divisor. But now we know that any common

divisor is, in fact, a divisor of their gtd.

Sometimes we need to do sums over all divisors of n. In this case it’s

often useful to use the handy rule

a, =

anlm,

m\n m\n

integer n > 0,

(4.7)

which holds since n/m runs through all divisors of n when m does. For

example, when n = 12 this says that al +

+ Q +

+ al2 = al2 +

al.

There’s also a slightly more general identity,

a, =

a,[n=mk],

m\n

k m>O

(4.8)

which is an immediate consequence of the definition (4.1). If n is positive, the

right-hand side of (4.8) is

tk,,,

on/k;

hence (4.8) implies (4.7). And equation

4.1 DIVISIBILITY 105

(4.8) works also when n is negative. (In such cases, the nonzero terms on the

right occur when k is the negative of a divisor of n.)

Moreover, a double sum over divisors can be “interchanged” by the law

ak,m =

ak,kl .

m\n

k\m

k\n

L\in/kl

(4.9)

For example, this law takes the following form when n = 12:

al,1 + (al.2 +

a2,2)

+ (al,3 +

a3,3)

+ fall4 +

a2,4

a4,4)

+ (al.6 +

a2,6

a3,6

a6,6)

+ tal,12 + a2,l2 +

a&12

a4,12

a6,12

a12,12)

tal.l

al.2

al.3

al.4 + al,6 +

al.12)

ta2,2

a2.4

a2,6

a&12) +

(a3,3

as,6

CQ12)

tad,4

q12)

(a6,6

a6,12)

a12,12.

We can prove (4.9) with Iversonian manipulation. The left-hand side is

ak.,[n=iml[m=kll

ak,kt[n=Ml;

i,l

k,m>O

k,1>0

the right-hand side is

ok.k~[n=jkl[n/k=mll

ak,kt[n=mlkl,

j,m

k,l>O

k.1>0

which is the same except for renaming the indices. This example indicates

that the techniques we’ve learned in Chapter 2 will come in handy as we study

number theory.

4.2 PRIMES

A positive integer p is called prime if it has just two divisors, namely

1 and

Throughout the rest of this chapter, the letter p will always stand

How about the p in

for a prime number, even when we don’t say so explicitly. By convention,

‘explicitly’?

1 isn’t prime, so the sequence of primes starts out like this:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, .,

Some numbers look prime but aren’t, like 91 (= 7.13) and 161 (= 7.23). These

numbers and others that have three or more divisors are called composite.

Every integer greater than 1 is either prime or composite, but not both.

Primes are of great importance, because they’re the fundamental building

blocks of all the positive integers. Any positive integer n can be written as a

106 NUMBER THEORY

product of primes,

n =

p,...pm

fiPk,

Pl 6

.‘.

6 Pm.

(4.10)

k=l

For example,

12=2.2.3;

11011

=7.11.11.13;

11111

=41.271.

(Products

denoted by

are analogous to sums denoted by

as explained in exer-

cise 2.25. If m = 0, we consider this to be an empty product, whose value

is 1 by definition; that’s the way n = 1 gets represented by (4.10).) Such a

factorization is always possible because if n > 1 is not prime it has a divisor

nl such that 1 < nl < n; thus we can write n = nl .nz, and (by induction)

we know that nl and

can be written as products of primes.

Moreover, the expansion in (4.10) is unique: There’s only one way to

write n as a product of primes in nondecreasing order. This statement is

called the Fundamental Theorem of Arithmetic, and it seems so obvious that

we might wonder why it needs to be proved. How could there be two different

sets of primes with the same product? Well, there can’t, but the reason isn’t

simply “by definition of prime numbers!’ For example, if we consider the set

of all real numbers of the form m +

when m and n are integers, the

product of any two such numbers is again of the same form, and we can call

such a number “prime” if it can’t be factored in a nontrivial way. The number

6 has two representations,

2.3

= (4 +

j(4

yet exercise 36 shows

that 2, 3, 4 +

and 4

are all “prime” in this system.

Therefore we should prove rigorously that (4.10) is unique. There is

certainly only one possibility when n = 1, since the product must be empty

in that case; so let’s suppose that n > 1 and that all smaller numbers factor

uniquely. Suppose we have two factorizations

n = p, . .

*Pm =

ql...qk,

Pl<...<Pm

and

ql<“‘<qk,

where the p’s and q’s are all prime. We will prove that

41.

If not, we

can assume that p, < q,, making p, smaller than all the q’s. Since p, and

are prime, their gcd must be 1; hence Euclid’s self-certifying algorithm gives

us integers a and b such that ap, + bql = 1. Therefore

q2..

qk +

b‘llqz...qk

qz...‘.jk.

Now p, divides both terms on the left, since

. . ,

= n; hence p, divides

the right-hand side, 42.. .

qk.

Thus 42.. . ok/p, is an integer, and 42.. .

has a prime factorization in which p, appears. But 42.. .

< n, so it has a

unique factorization (by induction). This contradiction shows that p, must

be equal to q, after all. Therefore we can divide both of n’s factorizations by

p,, obtaining

. . .p,,, = 42.. .

< n. The other factors must likewise be

equal (by induction), so our proof of uniqueness is complete.

4.2 PRIMES 107

It’s the factor-

Sometimes it’s more useful to state the Fundamental Theorem in another

ization, not the

theorem, that’s

way: Every positive integer can be written uniquely in the form

unique.

nP”Y

where each

3 0.

(4.11)

The right-hand side is a product over infinitely many primes; but for any

particular n all but a few exponents are zero, so the corresponding factors

are 1. Therefore it’s really a finite product, just as many “infinite” sums are

really finite because their terms are mostly zero.

Formula (4.11) represents n uniquely, so we can think of the sequence

(nz, n3, n5, . ) as a number system for positive integers. For example, the

prime-exponent representation of 12 is

(2,1,0,0,.

. . ) and the prime-exponent

representation of 18 is

(1,2,0,0,

To multiply two numbers, we simply

add their representations. In other words,

k = mn

m,+n,

forallp.

(4.12)

This implies that

m\n

and it follows immediately that

mp <

for all p,

(4.13)

k = gcd(m,n)

min(m,,n,)

for allp;

(4.14)

k = lcm(m,n)

max(m,,n,)

for all

(4.15)

For example, since 12 =

.3’ and 18 = 2’ .

32,

we can get their gcd and lcm

by taking the min and max of common exponents:

gcd(12,18) = 2

min(2,li

.3min(l,21 =

.31

= 6;

lcm(12,18) = 2

maX(2,1)

. 3max(l,2) =

.32 = 36.

If the prime p divides a product mn then it divides either m or n, perhaps

both, because of the unique factorization theorem. But composite numbers

do not have this property. For example, the nonprime 4 divides 60 = 6.10,

but it divides neither 6 nor 10. The reason is simple: In the factorization

60 =

6.10

(2.3)(2.5),

the two prime factors of 4 =

2.2

have been split

into two parts, hence 4 divides neither part. But a prime is unsplittable, so

it must divide one of the original factors.

4.3

PRIME EXAMPLES

How many primes are there? A lot. In fact, infinitely many. Euclid

proved this long ago in his Theorem 9: 20, as follows. Suppose there were

108 NUMBER THEORY

only finitely many primes, say k of

them--,

3, 5, . . . ,

Pk.

Then, said Euclid,

we should consider the number

M =

2’3’5’..:Pk

+ 1.

None of the k primes can divide M, because each divides M

1. Thus there

must be some other prime that divides M; perhaps M itself is prime. This

contradicts our assumption that 2, 3, . . . ,

are the only primes, so there

must indeed be infinitely many.

Euclid’s proof suggests that we define Euclid numbers by the recurrence

elez...e,-1

whenn>l.

(4.16)

The sequence starts out

=I+1

=2;

=2+1

=3;

= 2.3+1 = 7;

= 2.3.7+1 = 43;

these are all prime. But the next case,

5, is

1807

13.139. It

turns out that

= 3263443 is prime, while

547.607.1033.31051;

= 29881~67003~9119521~6212157481.

It is known that

es,

. . . ,

e17

are composite, and the remaining

are probably

composite as well. However, the Euclid numbers are all

reZatiweZy

prime to

each other; that is,

gcd(e,,e,)

when m # n.

Euclid’s algorithm (what else?) tells us this in three short steps, because

mod

= 1 when n > m:

gc4em,e,)

gcd(l,e,)

= gcd(O,l) =

Therefore, if we let

be the smallest factor of

for all j 3

the primes 41,

q2,

(73,

. . .

are all different. This is a sequence of infinitely many primes.

Let’s pause to consider the Euclid numbers from the standpoint of Chap-

ter 1. Can we express

in closed form? Recurrence (4.16) can be simplified

by removing the three dots: If n > 1 we have

cdot

7rpLjro1

lvpopoi

nkiov<

murb~

706

Xp0rE&ur0(

7rXijOOV~

7rphwu

IypLep(;Iu.~~

-Euclid [SO]

[Translation:

“There are

primes than in

any given set

of primes.

“1

en-2en-l +

(en-l

-l)e,-j

fl =

&,-qp,

4.3 PRIME EXAMPLES 109

Thus

has about twice as many decimal digits as e,-1 . Exercise 37 proves

that there’s a constant E

1.264

such that

(4.17)

And exercise 60 provides a similar formula that gives nothing but primes:

lp3"J

for some constant P. But equations like (4.17) and (4.18) cannot really be

considered to be in closed form, because the constants E and P are computed

from the numbers

and

p,,

in a sort of sneaky way. No independent re-

lation is known (or likely) that would connect them with other constants of

mathematical interest.

Indeed, nobody knows any useful formula that gives arbitrarily large

primes but only primes. Computer scientists at Chevron Geosciences did,

however, strike mathematical oil in 1984. Using a program developed by

David Slowinski, they discovered the largest prime known at that time,

2216091

-1

probably more,

by the time you

read

this.

while testing a new Cray X-MP supercomputer. It’s easy to compute this

number in a few milliseconds on a personal computer, because modern com-

puters work in binary notation and this number is simply (11 . .

.1)2.

All

216 091 of its bits are ‘1'. But it’s much harder to prove that this number

is prime. In fact, just about any computation with it takes a lot of time,

because it’s so large. For example, even a sophisticated algorithm requires

several minutes just to convert

22’609’

to radix

on a PC. When printed

out, its 65,050 decimal digits require 65 cents U.S. postage to mail first class.

Incidentally,

22’609’

1 is the number of moves necessary to solve the

Tower of Hanoi problem when there are 216,091 disks. Numbers of the form

(where p is prime, as always in this chapter) are called Mersenne numbers,

after Father Marin Mersenne who investigated some of their properties in the

seventeenth century. The Mersenne primes known to date occur for

p = 2, 3,

5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253,

4423,

9689,9941,

11213,19937,21701,

23209,44497,

86243,110503,

132049,

and 216091.

The number 2”

1 can’t possibly be prime if n is composite, because

2k”

has 2”’

1 as a

factor:

2km

(2"

l)(2mckp')

+2"'+2)

+...+

1).

110 NUMBER THEORY

But

1 isn’t always prime when p is prime;

2”

1 = 2047 = 23.89 is the

smallest such nonprime. (Mersenne knew this.)

Factoring and primality testing of large numbers are hot topics nowadays.

A summary of what was known up to 1981 appears in Section 4.5.4 of

[174],

and many new results continue to be discovered. Pages 391-394 of that book

explain a special way to test Mersenne numbers for primality.

For most of the last two hundred years, the largest known prime has

been a Mersenne prime, although only 31 Mersenne primes are known. Many

people are trying to find larger ones, but it’s getting tough. So those really

interested in fame (if not fortune) and a spot in The Guinness Book of World

Records might instead try numbers of the form 2nk +

for small values of k

like 3 or 5. These numbers can be tested for primality almost as quickly as

Mersenne numbers can; exercise 4.5.4-27 of

[174]

gives the details.

We haven’t fully answered our original question about how many primes

there are. There are infinitely many, but some infinite sets are “denser” than

others. For instance, among the positive integers there are infinitely many

even numbers and infinitely many perfect squares, yet in several important

senses there are more even numbers than perfect squares. One such sense

looks at the size of the nth value. The nth even integer is 2n and the nth

perfect square is

n’;

since 2n is much less than

for large n, the nth even

integer occurs much sooner than the nth perfect square, so we can say there

are many more even integers than perfect squares. A similar sense looks at

the number of values not exceeding x. There are 1x/2] such even integers and

L&j

perfect squares; since x/2 is much larger than

for large x, again we

can say there are many more even integers.

What can we say about the primes in these two senses? It turns out that

the nth prime, P,, is about n times the natural log of n:

pll

nlnn.

(The symbol

‘N’

can be read “is asymptotic to”; it means that the limit of

the ratio

PJnlnn

is 1 as n goes to infinity.) Similarly, for the number of

primes

n(x)

not exceeding x we have what’s known as the prime number

theorem:

Proving these two facts is beyond the scope of this book, although we can

show easily that each of them implies the other. In Chapter 9 we will discuss

the rates at which functions approach infinity, and we’ll see that the func-

tion nlnn, our approximation to

P,,

lies between 2n and

asymptotically.

Hence there are fewer primes than even integers, but there are more primes

than perfect squares.

Weird. I thought

there were the same

number of even

integers as per-

fect squares, since

there’s a one-to-one

correspondence

between them.

4.3 PRIME EXAMPLES 111

These formulas, which hold only in the limit as n or

03, can be

replaced by more exact estimates. For example, Rosser and Schoenfeld

[253]

have established the handy bounds

lnx-i

lnx-t,

for x 3 67;

(4.19)

n(lnn+lnlnn-3)

n(lnn+lnlnn-t),

forn320.

(4.20)

If we look at a “random” integer n, the chances of its being prime are

about one in Inn. For example, if we look at numbers near

1016,

we’ll have to

examine about 16 In 10 x 36.8 of them before finding a prime. (It turns out

that there are exactly 10 primes between

1016

370 and

1016

1.) Yet the

distribution of primes has many irregularities. For example, all the numbers

between

+ 2 and

. . .

P,+l

1 inclusive are composite.

Many examples of “twin primes” p and p + 2 are known (5 and 7, 11 and 13,

17and19,29and31,

. . . . 9999999999999641 and 9999999999999643, . . .

yet

nobody knows whether or not there are infinitely many pairs of twin primes.

(See Hardy and Wright

[150,

$1.4 and

52.81.)

One simple way to calculate all

X(X)

primes 6 x is to form the so-called

sieve of Eratosthenes: First write down all integers from 2 through x. Next

circle 2, marking it prime, and cross out all other multiples of 2. Then repeat-

edly circle the smallest uncircled, uncrossed number and cross out its other

multiples. When everything has been circled or crossed out, the circled num-

bers are the primes. For example when x = 10 we write down 2 through 10,

circle 2, then cross out its multiples 4, 6, 8, and 10. Next 3 is the smallest

uncircled, uncrossed number, so we circle it and cross out 6 and 9. Now

5 is smallest, so we circle it and cross out 10. Finally we circle 7. The circled

numbers are 2, 3, 5, and 7; so these are the

10) = 4 primes not exceeding 10.

“Je

sers

de la

z”;$

Zg$;/f

4.4

FACTORIAL FACTORS

produif

nombres

dkroissans

depuis

Now let’s take a look at the factorization of some interesting highly

jusqu9

l’unitk,

composite numbers, the factorials:

saioir-n(n

2). 3.2.1.

L’emploi continue/

l’analyse

combi-

natoire que je fais

dans /a plupart de

mes dCmonstrations,

rendu

cette

nota-

tion

indispensa

b/e.

”

Ch. Kramp (186]

n! =

1.2...:n

fib

integer n 3 0.

(4.21)

k=l

According to our convention for an empty product, this defines O! to be 1.

Thus n! = (n

1 )! n for every positive integer n. This is the number of

permutations of n distinct objects. That is, it’s the number of ways to arrange

n things in a row: There are n choices for the first thing; for each choice of

first thing, there are n

1 choices for the second; for each of these

n(n

choices, there are n

2 for the third; and so on, giving n(n

1) (n

2) . . . (1)

112 NUMBER THEORY

arrangements in all. Here are the first few values of the factorial function.

n 01234

5 6 7

8 9 10

1 1 2 6 24

120 720 5040 40320 362880 3628800

It’s useful to know a few factorial facts, like the first six or so values, and the

fact that

lo!

is about

million plus change; another interesting fact is that

the number of digits in n! exceeds n when n > 25.

We can prove that n! is plenty big by using something like Gauss’s trick

of Chapter 1:

n!’

= (1

.2...:n)(n...

:2.1) =

fik(n+l-k).

k=l

We have n 6 k(n + 1

(n + 1

)2,

since the quadratic polynomial

k(n+l

-k)

a(r~+l)~-

(k-

$(n+

1))2

has its smallest value at k = 1 and

its largest value at k =

(n + 1). Therefore

k=l k=l

that is,

n/2

(n+

l)n

(4.22)

This relation tells us that the factorial function grows exponentially!!

To approximate n! more accurately for large n we can use Stirling’s

formula, which we will derive in Chapter 9:

&Gi(:)n.

(4.23)

And a still more precise approximation tells us the asymptotic relative error:

Stirling’s formula undershoots n! by a factor of about 1

12n). Even for fairly

small n this more precise estimate is pretty good. For example, Stirling’s

approximation

(4.23)

gives a value near 3598696 when n = 10, and this is

about 0.83%

l/l20

too small. Good stuff,

asymptotics.

But let’s get back to primes. We’d like to determine, for any given

prime

the largest power of p that divides n!; that is, we want the exponent

of p in

n!‘s

unique factorization. We denote this number by

(n!), and we

start our investigations with the small case p = 2 and n = 10. Since

lo!

is the

product of ten numbers,

e:2(

lo!)

can be found by summing the powers-of-2

4.4 FACTORIAL FACTORS 113

contributions of those ten numbers; this calculation corresponds to summing

the columns of the following array:

23456789101powersof2

divisible by 2 x x x x x 5 =

[10/2J

divisible by 4

2 =

[10/4]

divisible by 8 X

1 = [10/S]

powersof

010201030 1 (

(The column sums form what’s sometimes called the ruler function p(k),

because of their similarity to

‘m

‘,

the lengths of lines marking

fractions of an inch.) The sum of these ten sums is 8; hence

divides

lo!

but

doesn’t.

There’s also another way: We can sum the contributions of the rows.

The first row marks the numbers that contribute a power of 2 (and thus are

divisible by 2); there are [10/2J = 5 of them. The second row marks those

that contribute an additional power of 2; there are

L10/4J

= 2 of them. And

the third row marks those that contribute yet another; there are

[10/S]

= 1 of

them. These account for all contributions, so we have

(1 O!) = 5 + 2 + 1 = 8.

For general n this method gives

ez(n!) =

This sum is actually finite, since the summand is zero when

> n. Therefore

it has only [lgn] nonzero terms, and it’s computationally quite easy. For

instance, when n = 100 we have

q(lOO!)

50+25+12+6+3+1

= 97.

Each term is just the floor of half the previous term. This is true for all n,

because as a special case of (3.11) we have lr~/2~+‘J = Lln/2k]

/2].

It’s espe-

cially easy to see what’s going on here when we write the numbers in binary:

100

= (1100100)~

=lOO

L100/2]

= (110010)~ = 50

L100/4] = (11001)2 = 25

1100/8] =

(1100)2

= 12

[100/16J

(110)2

= 6

1100/32]

(1l)z

= 3

[100/64J

(I)2

We merely drop the least significant bit from one term to get the next.

114 NUMBER THEORY

The binary representation also shows us how to derive another formula,

E~(TI!)

n-Y2(n)

(4.24)

where ~z(n) is the number of l’s in the binary representation of n. This

simplification works because each 1 that contributes 2”’ to the value of n

contributes 2”-’ +

2mP2

+ . .

.+2’=2”-1

tothevalueofcz(n!).

Generalizing our findings to an arbitrary prime p, we have

(4.25)

by the same reasoning as before.

About how large is c,(n!)? We get an easy (but good) upper bound by

simply removing the floor from the summand and then summing an infinite

geometric progression:

e,(n!) <

i+l+n+...

(

,+i+$+...

-n

-0

P-1

=p_l.

For p = 2 and n = 100 this inequality says that 97 < 100. Thus the up-

per bound 100 is not only correct, it’s also close to the true value 97. In

fact, the true value n

VI(~)

n in general, because ~z(n) 6 [lgnl is

asymptotically much smaller than n.

When p = 2 and 3 our formulas give

ez(n!)

n and

e3(n!)

n/2, so

it seems reasonable that every once in awhile

(n!) should be exactly half

as big as

ez(n!).

For example, this happens when n = 6 and n = 7, because

24.

7!/7.

But nobody has yet proved that such coincidences

happen infinitely often.

The bound on e,(n!) in turn gives us a bound on

p”~(~!),

which is p’s

contribution to n! :

Gin!)

pw(P-‘)

And we can simplify this formula (at the risk of greatly loosening the upper

bound) by noting that p <

2pP’;

hence

pn/(Pme’)

6 (2p-‘)n/(pp’) =

2”.

other words, the contribution that any prime makes to n! is less than 2”.

4.4 FACTORIAL FACTORS 115

We can use this observation to get another proof that there are infinitely

many primes. For if there were only the k primes 2, 3, . . . ,

Pk,

then we’d

have n! <

(2”)k

2nk

for all n

1, since each prime can contribute at most

a factor of 2”

1. But we can easily contradict the inequality n! <

2”k

choosing n large enough, say n =

22k.

Then

contradicting the inequality n! > n

n/2

that we derived in (4.22). There are

infinitely many primes, still.

We can even beef up this argument to get a crude bound on n(n), the

number of primes not exceeding n. Every such prime contributes a factor of

less than

2”

to n!; so, as before,

n! < 2nn(n).

If we replace n! here by Stirling’s approximation (4.23), which is a lower

bound, and take logarithms, we get

nrr(n) > nlg(n/e) +

lg(27rn)

;

hence

This lower bound is quite weak, compared with the actual value z(n)

n/inn,

because logn is much smaller than n/logn when n is large. But we

didn’t have to work very hard to get it, and a bound is a bound.

4.5

RELATIVE PRIMALITY

When gcd(m, n) =

the integers m and n have no prime factors in

common and we say that they’re relatively prime.

This concept is so important in practice, we ought to have a special

notation for it; but alas, number theorists haven’t come up with a very good

one yet. Therefore we cry: HEAR us, 0 MATHEMATICIANS OF THE WORLD!

LETUS

NOTWAITANYLONGER! WE CAN MAKEMANYFORMULAS CLEARER

Like perpendicular

BY DEFINING A NEW NOTATION NOW! LET us AGREE TO WRITE ‘m

n’,

lines don

‘t

have

m A ND n

ARE

RELATIVELY

PRIME.

a common direc-

AND TO SAY

IS PRIME TO

Tl.;

tion, perpendicular

In other words, let us declare that

numbers don’t have

common factors.

ml-n

m,n

are integers and gcd(m,n) =

(4.26)

116 NUMBER THEORY

A fraction m/n is in lowest terms if and only if m

n. Since we

reduce fractions to lowest terms by casting out the largest common factor of

numerator and denominator, we suspect that, in general,

mlgcd(m,n)

n/gcd(m,

;

(4.27)

and indeed this is true. It follows from a more general law, gcd(km, kn) =

kgcd(m, n), proved in exercise 14.

The

relation has a simple formulation when we work with the prime-

exponent representations of numbers, because of the gcd rule (4.14):

mln

min(m,,n,)

= 0 for allp.

(4.28)

Furthermore, since mP and

are nonnegative, we can rewrite this as

The dot product is

zero, like orthogonal

mln

mPnP

= 0 forallp.

(4.2g) vectors.

And now we can prove an important law by which we can split and combine

two

relations with the same left-hand side:

klm

and

kin

mn.

(4.30)

In view of (4.2g), this law is another way of saying that k,,mp = 0 and

kpnp = 0 if and only if

(mp +

np)

= 0, when mp and

are nonnegative.

There’s a beautiful way to construct the set of all nonnegative fractions

m/n with m

n, called the

Stem-Brocot

tree because it was discovered

Interesting how

independently by Moris Stern

[279],

a German mathematician, and Achille

mathematicians

Brocot

[35],

a French clockmaker. The idea is to start with the two fractions

will say

“discov-

(y ,

and then to repeat the following operation as many times as desired:

ered”

when

abso-

lute/y anyone e/se

would

have said

Insert

m+m’

between two adjacent fractions

and

The new fraction

(m+m’)/(n+n’)

is called the mediant of m/n and

m’/n’.

For example, the first step gives us one new entry between

and

and the next gives two more:

11 21

7, 23

7, 7,

The next gives four more,

011213231

4.5 RELATIVE PRIMALITY 117

and then we’ll get 8, 16, and so on. The entire array can be regarded as an

/guess l/O is

infinite binary tree structure whose top levels look like this:

infinity, “in lowest

terms.”

Each fraction is

where F is the nearest ancestor above and to the left,

and

is the nearest ancestor above and to the right. (An “ancestor” is a

fraction that’s reachable by following the branches upward.) Many patterns

can be observed in this tree.

Conserve parody.

Why does this construction work? Why, for example, does each mediant

fraction

(mt

m’)/(n +n’) turn out to be in lowest terms when it appears in

this tree? (If m, m’, n, and n’ were all odd, we’d get even/even; somehow the

construction guarantees that fractions with odd numerators and denominators

never appear next to each other.) And why do all possible fractions m/n occur

exactly once? Why can’t a particular fraction occur twice, or not at all?

All of these questions have amazingly simple answers, based on the fol-

lowing fundamental fact: If m/n and m//n’ are consecutive fractions at any

stage of the construction, we have

m’n-mn’ = 1.

(4.31)

This relation is true initially (1 . 1

0.0 = 1); and when we insert a new

mediant (m + m’)/(n + n’), the new cases that need to be checked are

(m+m’)n-m(n+n’)

= 1;

m’(n + n’)

(m + m’)n’ = 1 .

Both of these equations are equivalent to the original condition (4.31) that

they replace. Therefore (4.31) is invariant at all stages of the construction.

Furthermore, if m/n < m’/n’ and if all values are nonnegative, it’s easy

to verify that

m/n < (m-t

m’)/(n+n’)

< m’/n’.

118 NUMBER THEORY

A mediant fraction isn’t halfway between its progenitors, but it does lie some-

where in between. Therefore the construction preserves order, and we couldn’t

possibly get the same fraction in two different places.

True, but if you get

One question still remains. Can any positive fraction a/b with a

comPound

frac-

possibly be omitted? The answer is no, because we can confine the

construe-

ture you’d better go

see

doctor,

tion to the immediate neighborhood of a/b, and in this region the behavior

is easy to analyze: Initially we have

-7

<(;)<A=$,

where we put parentheses around

to indicate that it’s not really present

yet. Then if at some stage we have

the construction forms (m +

m’)/(n

+ n’) and there are three cases. Either

(m +

m’)/(n

+ n’) = a/b and we win; or (m +

m’)/(n

+ n’) < a/b and we

can set m +- m + m’, n +- n + n’; or (m +

m’)/(n

+ n’) > a/b and we

can set m’

m + m’, n’

n + n’. This process cannot go on indefinitely,

because the conditions

“-F

> 0

and

n’

;>o

imply that

an-bm 3 1 and bm’

an’ 3 1;

hence

(m’+n’)(an-bm)+(m+n)(bm’-an’)

m’+n’+m+n;

and this is the same as a + b 3 m’ + n’ + m + n by (4.31). Either m or n or

m’ or n’ increases at each step, so we must win after at most a + b steps.

The Farey series of order N, denoted by 3~, is the set of all reduced

fractions between 0 and 1 whose denominators are N or less, arranged in

increasing order. For example, if N = 6 we have

11112

1.3

1'6'5'4'3'5'2'5'3'4'5'6'1'

We can obtain 3~ in general by starting with

= 9,

and then inserting

mediants whenever it’s possible to do so without getting a denominator that

is too large. We don’t miss any fractions in this way, because we know that

the Stern-Brocot construction doesn’t miss any, and because a mediant with

denominator 6 N is never formed from a fraction whose denominator is > N.

(In other words, 3~ defines a subtree of the Stern-Brocot tree, obtained by

Fdrey ‘nough.

4.5 RELATIVE PRIMALITY 119

pruning off unwanted branches.) It follows that m’n

mn’ = 1 whenever

m/n and m//n’ are consecutive elements of a Farey series.

This method of construction reveals that

can be obtained in a simple

way from

3~~1:

We simply insert the fraction (m + m’)/N between con-

secutive fractions m/n, m//n’ of

3~~1

whose denominators sum to N. For

example, it’s easy to obtain

from the elements of

36,

by inserting f ,

. . . , f according to the stated rule:

3, =

111

112

1'7'6'5'4'7'3'5'7'2'7'5'3'7'4'5'6'7'1'

When N is prime, N

1 new fractions will appear; but otherwise we’ll have

fewer than N

1, because this process generates only numerators that are

relatively prime to N.

Long ago in (4.5) we proved-in different words-that whenever m

and 0 < m 6 n we can find integers a and b such that

ma-nb =

(4.32)

(Actually we said m’m + n’n = gcd( m,

n),

but we can write 1 for gcd( m,

n),

a for m’, and b for -n’.) The Farey series gives us another proof of

(4.32),

because we can let b/a be the fraction that precedes m/n in

3,,.

Thus (4.5)

is just (4.31) again. For example, one solution to 3a

7b = 1 is a = 5, b = 2,

since

precedes

37.

This construction implies that we can always find a

solution to (4.32) with 0 6 b < a < n, if 0 < m < n. Similarly, if 0 6 n < m

and m

n, we can solve (4.32) with 0 < a 6 b 6 m by letting a/b be the

fraction that follows n/m in

3m.

Sequences of three consecutive terms in a Farey series have an amazing

property that is proved in exercise 61. But we had better not discuss the

Farey series any further, because the entire Stern-Brocot tree turns out to be

even more interesting.

We can, in fact, regard the Stern-Brocot tree as a number system for

representing rational numbers, because each positive, reduced fraction occurs

exactly once. Let’s use the letters L and R to stand for going down to the

left or right branch as we proceed from the root of the tree to a particular

fraction; then a string of L’s and R’s uniquely identifies a place in the tree.

For example, LRRL means that we go left from f down to

then right to

then left to

We can consider LRRL to be a representation

Every positive fraction gets represented in this way as a unique string

of L’s and R’s.

Well, actually there’s a slight problem: The fraction f corresponds to

the empty string, and we need a notation for that. Let’s agree to call it I,

because that looks something like 1 and it stands for “identity!’

120 NUMBER THEORY

This representation raises two natural questions: (1) Given positive inte-

gers m and n with m

n, what is the string of L’s and R’s that corresponds

to m/n? (2) Given a string of L’s and R’S, what fraction corresponds to it?

Question 2 seems easier, so let’s work on it first. We define

f(S) = fraction corresponding to S

when S is a string of L’s and R’s. For example, f (LRRL) =

According to the construction, f(S) = (m + m’)/(n + n’) if m/n and

m’/n’ are the closest fractions preceding and following S in the upper levels

of the tree. Initially m/n = O/l and m’/n’ = l/O; then we successively

replace either m/n or m//n’ by the mediant (m + m’)/(n + n’) as we move

right or left in the tree, respectively.

How can we capture this behavior in mathematical formulas that are

easy to deal with? A bit of experimentation suggests that the best way is to

maintain a 2 x 2 matrix

that holds the four quantities involved in the ancestral fractions m/n and

m//n’ enclosing S. We could put the m’s on top and the n’s on the bottom,

fractionwise; but this upside-down arrangement works out more nicely be-

cause we have M(1) = (A:) when the process starts, and (A!) is traditionally

called the identity matrix I.

A step to the left replaces n’ by n + n’ and m’ by m + m’; hence

(This is a special case of the general rule

for multiplying 2 x 2 matrices.) Similarly it turns out that

M(SR) =

;;;,

;,)

= W-9

(;

;)

Therefore if we define L and R as 2 x 2 matrices,

If you’re

clueless

about matrices,

don’t panic; this

book uses them

only here.

(4.33)

4.5 RELATIVE PRIMALITY 121

we get the simple formula M(S) = S, by induction on the length of S. Isn’t

that nice? (The letters L and R serve dual roles, as matrices and as letters in

the string representation.) For example,

M(LRRL)

LRRL

(;;)(;:)(;$(;;)

(f;)(;;)

(ii);

the ancestral fractions that enclose LRRL =

are

and

And this con-

struction gives us the answer to Question 2:

f(S) =

f((L

Z,))

(4.34)

How about Question

That’s easy, now that we understand the fun-

damental connection between tree nodes and 2 x 2 matrices. Given a pair of

positive integers m and n, with m

n, we can find the position of m/n in

the Stern-Brocot tree by “binary search” as follows:

s := I;

while m/n # f(S) do

if m/n < f(S) then (output(L); S := SL)

else (output(R); S := SR)

This outputs the desired string of L’s and R’s.

There’s also another way to do the same job, by changing m and n instead

of maintaining the state S. If S is any 2 x 2 matrix, we have

f(RS) = f(S)+1

because RS is like S but with the top row added to the bottom row. (Let’s

look at it in slow motion:

n’

m+n

m’fn’

hence f(S) =

(m+m’)/(n+n’)

and f(RS) = ((m+n)+(m’+n’))/(n+n’).)

If we carry out the binary search algorithm on a fraction m/n with m > n,

the first output will be R; hence the subsequent behavior of the algorithm will

have f(S) exactly 1 greater than if we had begun with (m

n)/n

instead of

m/n. A similar property holds for L, and we have

= f(RS)

m-n

= f(S)) when m > n;

n n

= f(LS)

= f(S)) when m <

n-m

122 NUMBER THEORY

This means that we can transform the binary search algorithm to the following

matrix-free procedure:

while m # n do

if m < n then (output(L); n := n-m)

else (output(R); m := m-n) .

For example, given m/n = 5/7, we have successively

m=5

1 1

n=7

2 2

output L R R L

in the simplified algorithm.

Irrational numbers don’t appear in the Stern-Brocot tree, but all the

rational numbers that are “close” to them do. For example, if we try the

binary search algorithm with the number e = 2.71828. . , instead of with a

fraction m/n, we’ll get an infinite string of L’s and R's that begins

RRLRRLRLLLLRLRRRRRRLRLLLLLLLLRLR....

We can consider this infinite string to be the representation of e in the Stern-

Brocot number system, just as we can represent e as an infinite decimal

2.718281828459...

or as an infinite binary fraction

(10.101101111110...)~.

Incidentally, it turns out that e’s representation has a regular pattern in the

Stern-Brocot system:

e = RL”RLRZLRL4RLR6LRL8RLR10LRL’2RL . . .

;

this is equivalent to a special case of something that Euler

[84]

discovered

when he was 24 years old.

From this representation we can deduce that the fractions

RRLRRLRLLLL R L R R R R R R

1 2 1 5 & 11 19 30 49 68 87

-------- 106 193 299 492 685 878 1071 1264

1'1'1'2'3' 4' 7'11'18'25'32' 39' 71'110'181'252'323' 394' 465""

are the simplest rational upper and lower approximations to e. For if m/n

does not appear in this list, then some fraction in this list whose numerator

is 6 m and whose denominator is < n lies between m/n and e. For example,

is not as simple an approximation as

= 2.714. . . , which appears in

the list and is closer to e. We can see this because the Stern-Brocot tree

not only includes all rationals, it includes them in order, and because all

fractions with small numerator and denominator appear above all less simple

ones. Thus,

= RRLRRLL is less than

= RRLRRL, which is less than

“Numerorum

congruentiam

hoc

signo,

posterum

deno-

tabimus, modulum

ubi opus erit in

clausulis adiun-

gentes,

-16

(mod. 5), -7 =

15 (mod.

ll).”

-C.

F. Gauss

11151

4.5 RELATIVE PRIMALITY 123

e = RRLRRLR.... Excellent approximations can be found in this way. For

example,

2.718280 agrees with e to six decimal places; we obtained this

fraction from the first 19 letters of e’s Stern-Brocot representation, and the

accuracy is about what we would get with 19 bits of e’s binary representation.

We can find the infinite representation of an irrational number

by a

simple modification of the matrix-free binary search procedure:

< 1 then (output(L);

au/(1

-K))

else (output(R);

(x-

1) .

(These steps are to be repeated infinitely many times, or until we get tired.)

If a is rational, the infinite representation obtained in this way is the same as

before but with RLm appended at the right of

01’s

(finite) representation. For

example, if 01=

we get RLLL . . . ,

corresponding to the infinite sequence of

fractions

,’

2’

3’

4’

*..I

which approach 1 in the limit. This situation is

exactly analogous to ordinary binary notation, if we think of L as 0 and R as 1:

Just as every real number x in

[O,

1) has an infinite binary representation

(.b,bZb3.. .

not ending with all l’s, every real number

[O,

00) has

an infinite Stern-Brocot representation

B2B3

. . . not ending with all R’s.

Thus we have a one-to-one order-preserving correspondence between [0, 1)

and [0, co) if we let 0

L and 1

There’s an intimate relationship between Euclid’s algorithm and the

Stern-Brocot representations of rationals. Given

= m/n, we get Lm/nJ

R’s, then

[n/(m

mod n)] L’s, then [(m mod

n)/(n

mod (m mod

n))]

R’s,

and so on. These numbers m mod n, n mod (m mod n), . . . are just the val-

ues examined in Euclid’s algorithm. (A little fudging is needed at the end

to make sure that there aren’t infinitely many R’s.) We will explore this

relationship further in Chapter 6.

4.6

‘MOD’: THE CONGRUENCE RELATION

Modular arithmetic is one of the main tools provided by number

theory. We got a glimpse of it in Chapter 3 when we used the binary operation

‘mod’, usually as one operation amidst others in an expression. In this chapter

we will use ‘mod’ also with entire equations, for which a slightly different

notation is more convenient:

b (mod m)

amodm = bmodm.

(4.35)

For example, 9 = -16 (mod 5), because 9 mod 5 = 4 = (-16) mod 5. The

formula ‘a = b (mod m)’ can be read “a is congruent to b modulo ml’ The

definition makes sense when a, b, and m are arbitrary real numbers, but we

almost always use it with integers only.

124 NUMBER THEORY

Since x mod m differs from x by a multiple of m, we can understand

congruences in another way:

b (mod m)

b is a multiple of m.

(4.36)

For if a mod m = b mod m, then the definition of ‘mod’ in (3.21) tells us

that a

b = a mod m + km

(b mod m +

Im)

= (k

l)m for some integers

k and

Conversely if a

b = km, then a = b if m = 0; otherwise

a mod m = a

[a/m]m = b + km

L(b

km)/mjm

b-[b/mJm

= bmodm.

The characterization of = in (4.36) is often easier to apply than (4.35). For

example, we have 8

23 (mod 5) because 8

23 = -15 is a multiple of 5; we

don’t have to compute both 8 mod 5 and 23 mod 5.

The congruence sign

‘

’

looks conveniently like

’

‘,

because

congru-

“I fee/

fine

today

ences

are almost like equations. For example, congruence is an equivalence

modulo a slight

relation; that is, it satisfies the reflexive law ‘a = a’, the symmetric law

headache.”

The Hacker’s

‘a

3 b

a’, and the transitive law ‘a

c’.

All these properties are easy to prove, because any relation

‘E’

that satisfies

‘a

c--J

f(a) = f(b)’ for some function f is an equivalence relation. (In

our case, f(x) = x mod m.) Moreover, we can add and subtract congruent

elements without losing congruence:

Dictionary

12771

a=b

and

c=d

a+c

b+d

(mod

;

a=b

and

c=d

===+

a-c

b-d

(mod m) .

For if a

b and c

d are both multiples of m, so are (a + c)

(b + d) =

b) + (c

d) and (a

d) = (a -b)

d). Incidentally, it

isn’t necessary to write ‘(mod m)’ once for every appearance of

‘

‘;

if the

modulus is constant, we need to name it only once in order to establish the

context. This is one of the great conveniences of congruence notation.

Multiplication works too, provided that we are dealing with integers:

b and c = d

(mod

integers b, c.

Proof: ac

bd = (a

b)c + b(c

d). Repeated application of this multipli-

cation property now allows us to take powers:

a-b

a”

b”

(mod

ml,

integers a, b;

integer n 3 0.

4.6 ‘MOD’: THE CONGRUENCE RELATION 125

For example, since 2

-1 (mod

3),

we have

(-1)” (mod 3); this means

that 2”

1 is a multiple of 3 if and only if n is even.

Thus, most of the algebraic operations that we customarily do with equa-

tions can also be done with congruences. Most, but not all. The operation

of division is conspicuously absent. If ad

bd (mod m), we can’t always

conclude that a

b. For example, 3.2

5.2 (mod

4),

but 3 8 5.

We can salvage the cancellation property for congruences, however, in

the common case that d and m are relatively prime:

ad=bd

a=b

(mod

(4.37)

integers a, b, d, m and d

For example, it’s

legit

to conclude from 15 E 35 (mod m) that 3 E 7 (mod m),

unless the modulus m is a multiple of 5.

To prove this property, we use the extended gcd law (4.5) again, finding

d’ and m’ such that d’d + m’m = 1. Then if ad

bd we can multiply

both sides of the congruence by d’, obtaining ad’d

bd’d. Since d’d

we have ad’d

a and bd’d

b; hence a

b. This proof shows that the

number d’ acts almost like l/d when congruences are considered (mod m);

therefore we call it the “inverse of d modulo m!’

Another way to apply division to congruences is to divide the modulus

as well as the other numbers:

ad = bd (modmd)

+=+

a = b (modm),

ford#O.

(4.38)

This law holds for all real a, b, d, and m, because it depends only on the

distributive law (a mod m) d = ad mod md: We have a mod m = b mod m

(a mod m)d = (b mod m)d

ad mod md = bd mod md. Thus,

for example, from 3.2

5.2 (mod 4) we conclude that 3

5 (mod 2).

We can combine (4.37) and (4.38) to get a general law that changes the

modulus as little as possible:

bd (mod m)

a=b

(

mod

gcd(d,

’

integers a, b, d, m.

(4.39)

For we can multiply ad

bd by d’, where d’d+ m’m = gcd( d, m); this gives

the congruence

gcd( d, m)

b. gcd( d, m) (mod m), which can be divided

gc44

ml.

Let’s look a bit further into this idea of changing the modulus. If we

know that a 3 b (mod loo), then we also must have a

b (mod lo), or

modulo any divisor of 100. It’s stronger to say that a

b is a multiple of 100

126 NUMBER THEORY

than to say that it’s a multiple of 10. In general,

a E b (mod md)

a = b (mod m) , integer d,

(4.40)

because any multiple of md is a multiple of m.

Conversely, if we know that a

‘=

b with respect to two small moduli, can

Modulitos?

we conclude that a E b with respect to a larger one? Yes; the rule is

a E b (mod m) and a

b (mod n)

a=b

(mod

lcm(m,

n)) , integers m, n > 0.

(4.41)

For example, if we know that a

b modulo 12 and 18, we can safely conclude

that a = b (mod 36). The reason is that if a

b is a common multiple of m

and n, it is a multiple of lcm( m, n). This follows from the principle of unique

factorization.

The special case m

n of this law is extremely important, because

lcm(m,

n) = mn when m and n are relatively prime. Therefore we will state

it explicitly:

a E b (mod mn)

a-b

(mod m) and a = b (mod n), if

min.

(4.42)

For example, a E b (mod 100) if and only if a E b (mod 25) and a E b

(mod 4). Saying this another way, if we know

mod 25 and x mod 4, then

we have enough facts to determine x mod 100. This is a special case of the

Chinese Remainder Theorem (see exercise 30), so called because it was

discovered by Sun Tsfi in China, about A.D. 350.

The moduli m and n in (4.42) can be further decomposed into relatively

prime factors until every distinct prime has been isolated. Therefore

a=b(modm)

arb(modp”p)

forallp,

if the prime factorization (4.11) of m is

pm”. Congruences modulo powers

of primes are the building blocks for all congruences modulo integers.

4.7 INDEPENDENT RESIDUES

One of the important applications of congruences is a residue num-

ber system, in which an integer x is represented as a sequence of residues (or

remainders) with respect to moduli that are prime to each other:

Res(x) = (x mod ml,. . .

mod m,) , if

mk for 1 6 j < k 6 r.

Knowing x mod

ml,

. . . , x mod

doesn’t tell us everything about x. But

it does allow us to determine x mod m, where m is the product

. . . m,.

4.7 INDEPENDENT RESIDUES 127

In practical applications we’ll often know that x lies in a certain range; then

we’ll know everything about x if we know

mod m and if m is large enough.

For example, let’s look at a small case of a residue number system that

has only two moduli, 3 and 5:

x mod 15

cmod3 (mod5

0 0 0

1 1

5 2

9 0

12 0

Each ordered pair (x mod 3, x mod 5) is different, because x mod 3 = y mod 3

andxmod5=ymod5ifandonlyifxmod15=ymod15.

We can perform addition, subtraction, and multiplication on the two

components independently, because of the rules of congruences. For example,

if we want to multiply 7 =

(1,2)

by 13 =

(1,3)

modulo 15, we calculate

l.lmod3=1and2.3mod5=1.

Theansweris(l,l)=l;hence7.13mod15

must equal 1. Sure enough, it does.

For

example,

the

Mersenne prime

23'-l

works well.

This independence principle is useful in computer applications, because

different components can be worked on separately (for example, by differ-

ent computers).

each modulus

is a distinct prime

pk,

chosen to be

slightly less than

23’,

then a computer whose basic arithmetic operations

handle integers in the range

L-2

3’

23’)

can easily compute sums, differences,

and products modulo pk. A set of r such primes makes it possible to add,

subtract, and multiply “multiple-precision numbers” of up to almost 31 r bits,

and the residue system makes it possible to do this faster than if such large

numbers were added, subtracted, or multiplied in other ways.

We can even do division, in appropriate circumstances. For example,

suppose we want to compute the exact value of a large determinant of integers.

The result will be an integer D, and bounds on

ID/

can be given based on the

size of its entries. But the only fast ways known for calculating determinants

128 NUMBER THEORY

require division, and this leads to fractions (and loss of accuracy, if we resort

to binary approximations). The remedy is to evaluate D mod

Dk,

for

VSIiOUS large primes

pk.

We can safely divide module

unless the divisor

happens to be a multiple of pk. That’s very unlikely, but if it does happen we

can choose another prime. Finally, knowing

for sufficiently many primes,

we’ll have enough information to determine D.

But we haven’t explained how to get from a given sequence of residues

(x mod

ml,

. . . ,x

mod m,) back to x mod m. We’ve shown that this conver-

sion can be done in principle, but the calculations might be so formidable

that they might rule out the idea in practice. Fortunately, there is a rea-

sonably simple way to do the job, and we can illustrate it in the situation

(x mod 3,x mod 5) shown in our little table. The key idea is to solve the

problem in the two cases

(1,O)

and (0,l); for if

(1,O)

= a and (0,l) = b, then

(x, y) = (ax + by) mod 15, since congruences can be multiplied and added.

In our case a = 10 and b = 6, by inspection of the table; but how could

we find a and b when the moduli are huge? In other words, if m

n, what

is a good way to find numbers a and b such that the equations

amodm = 1, amodn = 0, bmodm = 0, bmodn = 1

all hold? Once again, (4.5) comes to the rescue: With Euclid’s algorithm, we

can find m’ and n’ such that

m’m+n’n = 1.

Therefore we can take a = n’n and b = m’m, reducing them both mod mn

if desired.

Further tricks are needed in order to minimize the calculations when the

moduli are large; the details are beyond the scope of this book, but they can

be found in

[174,

page

2741.

Conversion from residues to the corresponding

original numbers is feasible, but it is sufficiently slow that we save total time

only if a sequence of operations can all be done in the residue number system

before converting back.

Let’s firm up these congruence ideas by trying to solve a little problem:

How many solutions are there to the congruence

1 (mod m) ,

(4.43)

if we consider two solutions x and x’ to be the same when x = x’?

According to the general principles explained earlier, we should consider

first the case that m is a prime power,

pk,

where k > 0. Then the congruence

= 1 can be written

(x-1)(x+1) = 0 (modpk),

4.7 INDEPENDENT RESIDUES 129

so p must divide either x

1 or x + 1, or both. But p can’t divide both

1 and x + 1 unless p = 2; we’ll leave that case for later. If p > 2, then

pk\(x

1)(x + 1)

pk\(x

1) or pk\(x + 1); so there are exactly two

solutions, x =

and x = -1.

The case p = 2 is a little different. If 2k\(~

1 )(x + 1) then either x

or x + 1 is divisible by 2 but not by 4, so the other one must be divisible

by 2kP’. This means that we have four solutions when k 3 3, namely x = *l

and x = 2k-’

1. (For example, when

= 8 the four solutions are x

5, 7 (mod 8); it’s often useful to know that the square of any odd integer has

the form 8n + 1.)

All primes are odd

except 2, which is

the oddest of all.

Now x2 = 1 (mod m) if and only if x2 = 1 (mod

pm”

) for all primes p

with mP > 0 in the complete factorization of m. Each prime is independent

of the others, and there are exactly two possibilities for x mod

pm”

except

when p = 2. Therefore if

has exactly

different prime divisors, the total

number of solutions to x2 = 1 is 2’, except for a correction when m. is even.

The exact number in general is

2~+[8\ml+[4\ml-[Z\ml

(4.44)

For example, there are four “square roots of unity modulo 12,” namely

7, and 11. When m = 15 the four are those whose residues mod 3 and mod 5

are

fl,

namely (1,

l),

(1,4), (2,

l),

and (2,4) in the residue number system.

These solutions are 1, 4,

11,

and 14 in the ordinary (decimal) number system.

4.8

ADDITIONAL APPLICATIONS

There’s some unfinished business left over from Chapter 3: We wish

to prove that the m numbers

Omodm, nmodm,

2nmodm,

. . . . (m-1)nmodm

(4.45)

consist of precisely d copies of the m/d numbers

d, 2d, . . . . m-d

in some order, where d = gcd(m, n). For example, when m = 12 and n = 8

we have d = 4, and the numbers are 0, 8, 4, 0, 8, 4, 0, 8, 4, 0, 8, 4.

The first part of the proof-to show that we get d copies of the first

Mathematicians love

m/d values-is now trivial. We have

to say that things

are trivial.

jn = kn (mod m)

j(n/d)

k(n/d) (mod m/d)

by (4.38); hence we get d copies of the values that occur when 0 6 k < m/d.

130 NUMBER THEORY

Now we must show that those m/d numbers are

(0,

d,2d,. . . , m

in some order. Let’s write m = m’d and n = n’d. Then kn mod m =

d(kn’ mod m’), by the distributive law (3.23); so the values that occur when

0 6 k < m’ are d times the numbers

0 mod m’, n’ mod m’, 2n’ mod m’, . . . ,

(m’

1 )n’ mod m’ .

But we know that m’

n’ by (4.27); we’ve divided out their gtd. Therefore

we need only consider the case d =

namely the case that m and n are

relatively prime.

So let’s assume that m

n. In this case it’s easy to see that the numbers

(4.45) are just

{O,

1, . . . , m

}

in some order, by using the “pigeonhole

principle!’ This principle states that if m pigeons are put into m pigeonholes,

there is an empty hole if and only if there’s a hole with more than one pigeon.

(Dirichlet’s box principle, proved in exercise 3.8, is similar.) We know that

the numbers (4.45) are distinct, because

kn (mod m)

k (mod m)

when m

n; this is (4.37). Therefore the m different numbers must fill all the

pigeonholes 0,

. . . ,

1. Therefore the unfinished business of Chapter 3

is finished.

The proof is complete, but we can prove even more if we use a direct

method instead of relying on the indirect pigeonhole argument. If m

n and

if a value j

[0, m) is given, we can explicitly compute k

[O,

m) such that

kn mod m = j by solving the congruence

j (mod m)

for k. We simply multiply both sides by n’, where m’m + n’n = 1, to get

jn’ [mod m)

;

hence k = jn’ mod m.

We can use the facts just proved to establish an important result discov-

ered by Pierre de Fermat in 1640. Fermat was a great mathematician who

contributed to the discovery of calculus and many other parts of mathematics.

He left notebooks containing dozens of theorems stated without proof, and

each of those theorems has subsequently been verified-except one. The one

that remains, now called “Fermat’s Last Theorem,” states that

a” + b” #

c”

(4.46)

4.8 ADDITIONAL APPLICATIONS 131

(NEWSFLASH]

Euler

1931

con-

jectured that

d4,

but

Noam

Elkies

found infinitely

many solutions in

August, 1987.

Now Roger Frye has

done an exhaustive

computer search,

proving (aRer about

I19 hours on a Con-

nection Machine)

that the smallest

solution is:

958004

+2175194

+4145604

4224814.

‘I.

laquelfe

propo-

sition, si

efle

est

vraie, est de

t&s

grand usage.”

-P. de Fermat

1971

for all positive integers a, b, c, and n, when n > 2. (Of course there are lots

of solutions to the equations a + b = c and

c2.)

This conjecture

has been verified for all n 6 150000 by Tanner and

Wagstaff

[285].

Fermat’s theorem of 1640 is one of the many that turned out to be prov-

able. It’s now called Fermat’s Little Theorem (or just Fermat’s theorem, for

short), and it states that

np-’ = 1 (modp),

ifnIp.

(4.47)

Proof: As usual, we assume that p denotes a prime. We know that the

p-l

numbersnmodp,2nmodp,

. . . . (p

1 )n mod p are the numbers

.“,

1 in some order. Therefore if we multiply them together we get

(2n). . . . . ((p

1)n)

(n mod p) . (2n mod p) . . . . . ((p

1)n mod p)

(p-l)!,

where the congruence is modulo p. This means that

l)!nP-’ =

(p-l)!

(modp),

and we can cancel the (p

l)!

since it’s not divisible by p. QED.

An alternative form of Fermat’s theorem is sometimes more convenient:

= n

(mod

) ,

integer n.

(4.48)

This congruence holds for all integers n. The proof is easy: If n

p we

simply multiply (4.47) by n. If not, p\n, so

3 0

In the same year that he discovered

(4.47),

Fermat wrote a letter to

Mersenne, saying he suspected that the number

22"

would turn out to be prime for all n 3 0. He knew that the first five cases

gave primes:

2'+1

= 3;

2'+1

= 5;

24+1

= 17;

28+1

= 257;

216+1

= 65537;

but he couldn’t see how to prove that the next case,

232

+ 1

4294967297,

would be prime.

It’s interesting to note that Fermat could have proved that

232

+ 1 is not

prime, using his own recently discovered theorem, if he had taken time to

perform a few dozen multiplications: We can set n = 3 in

(4.47),

deducing

that

p3’

(mod

232

+ l), if

232

+ 1 is prime.

132 NUMBER THEORY

And it’s possible to test

this,

relation by hand, beginning with 3 and squaring

32 times, keeping only the remainders mod

232

+ 1. First we have

= 9,

this is

Fermat’s

then

32;’

= 81, then

323

= 6561, and so on until we reach

32"

3029026160

(mod

232

+ 1) .

Little Theorem,

the other one was

last

but not least.

The result isn’t 1, so

232

+ 1 isn’t prime. This method of disproof gives us

no clue about what the factors might be, but it does prove that factors exist.

(They are 641 and 6700417.)

3232

had turned out to be

modulo

232

+ 1, the calculation wouldn’t

have proved that

232

+ 1 is prime; it just wouldn’t have disproved it. But

exercise 47 discusses a converse to Fermat’s theorem by which we can prove

that large prime numbers are prime, without doing an enormous amount of

laborious arithmetic.

We proved Fermat’s theorem by cancelling (p

1 )! from both sides of a

congruence. It turns out that (p

I)! is always congruent to -1, modulo p;

this is part of a classical result known as Wilson’s theorem:

(n-- I)! 3 -1 (mod n) n is prime,

ifn>l.

(4.49)

One half of this theorem is trivial: If n > 1 is not prime, it has a prime

divisor p that appears as a factor of (n

l)!, so (n

l)!

cannot be congruent

to -1. (If (n- 1 )! were congruent to -1 modulo n, it would also be congruent

to -1 modulo p, but it isn’t.)

The other half of Wilso’n’s theorem states that (p

l)!

-1 (mod p).

We can prove this half by p,airing up numbers with their inverses mod p. If

p, we know that there exists n’ such that

n’n

(mod

P);

here n’ is the inverse of n, and n is also the inverse of n’. Any two inverses

of n must be congruent to each other, since nn’ E nn” implies n’

n”.

p is

prime, is

Now suppose we pair up each number between 1 and p-l with its inverse.

prime

prime?

Since the product of a number and its inverse is congruent to

the product

of all the numbers in all pairs of inverses is also congruent to 1; so it seems

that (p -- l)! is congruent to 1. Let’s check, say for p = 5. We get

= 24;

but this is congruent to 4, not

modulo 5. Oops- what went wrong? Let’s

take a closer look at the inverses:

1’

2' = 3, 3' = 2,

4' = 4.

Ah so; 2 and 3 pair up but 1 and 4 don’t-they’re their own inverses.

To resurrect our analysis we must determine which numbers are their

own inverses. If x is its own inverse, then

= 1 (mod p); and we have

4.8 ADDITIONAL APPLICATIONS 133

“5

fuerit

N ad x

numerus

primus

et n numerus

partium ad N

primarum,

turn

potestas

unitate

minuta

semper per

numerum

erit

divisibilis.”

-L.

Euler

[89]

already proved that this congruence has exactly two roots when p > 2. (If

p = 2 it’s obvious that (p

l)!

= -1, so we needn’t worry about that case.)

The roots are

and

and the other numbers (between

and

pair

up; hence

(p-l)!

l.(p-1)

-1,

as desired.

Unfortunately, we can’t compute factorials efficiently, so Wilson’s theo-

rem is of no use as a practical test for primality. It’s just a theorem.

4.9 PHI AND MU

How many of the integers

(0,

1, . . . , m-l} are relatively prime to m?

This is an important quantity called cp(m), the “totient” of m (so named by

J. J. Sylvester

[284],

a British mathematician who liked to invent new words).

We have

q(l)

= 1,

q(p)

= p

1, and cp(m) < m- 1 for all composite

numbers m.

The

function is called Euler’s totient j’unction, because Euler was the

first person to study it. Euler discovered, for example, that Fermat’s theorem

(4.47) can be generalized to nonprime moduli in the following way:

nVp(m)

= 1 (mod m) ,

ifnIm.

(4.50)

(Exercise 32 asks for a proof of Euler’s theorem.)

If m is a prime power

pk,

it’s easy to compute cp(m), because n

p%n. The multiples of

{O,l,...,pk

-l} are

{0,p,2p,...,pk

-p};

hence

there are

pk-'

of them, and

cp(pk)

counts what is left:

cp(pk)

= pk

pk-’

Notice that this formula properly gives

q(p)

= p

when k =

If m > 1 is not a prime power, we can write m = ml rn2 where ml

m2.

Then the numbers 0 6 n < m can be represented in a residue number system

as (n mod

ml,

n mod ml). We have

nlm

nmodml

ml and nmod ml

rn2

by (4.30) and (4.4). Hence, n mod m is “good” if and only if n mod ml

and n mod rn2 are both “good,”

if we consider relative primality to be a

virtue. The total number of good values modulo m can now be computed,

recursively: It is q(rnl )cp(mz), because there are cp(ml ) good ways to choose

the first component n mod ml and cp(m2) good ways to choose the second

component n mod rn2 in the residue representation.

134 NUMBER THEORY

For example, (~(12) =

cp(4)(p(3)

= 292 = 4, because n is prime to 12 if

“Sisint

A et B

nu-

and only if n mod 4 = (1 or

and n mod 3 = (1 or 2). The four values prime

meri inter se primi

to 12 are

(l,l),

(1,2),

(3,111,

(3,2) in the residue number system; they are

et numerus partium

1, 5, 7, 11 in ordinary decimal notation. Euler’s theorem states that

3 1

primarum

sjt

= a,

numerus

(mod 12) whenever n

12.

vero partium ad B

A function f(m) of positive integers is called

mult$icative

if f (1) = 1 ~~f~u~e$

raz’

and

tium ad productum

primarum

erit

f(mlm2)

f(m)f(m2)

whenever

mz.

(4’5l) =

“‘:L.

Euler

[#J]

We have just proved that q)(m) is multiplicative. We’ve also seen another

instance of a multiplicative function earlier in this chapter: The number of

incongruent solutions to x’ = 1 (mod m) is multiplicative. Still another_

example is f(m) =

for any power

01.

A multiplicative function is defined completely by its values at prime

powers, because we can decompose any positive integer m into its prime-

power factors, which are relatively prime to each other. The general formula

f(m)

nf(pmpl,

if m=

pmP

(4.52)

P P

holds if and only if f is multiplicative.

In particular, this formula gives us the value of Euler’s totient function

for general m:

q(m)

n(p”p

-pm,-‘)

mn(l

-J-).

P\m P\m

For example, (~(12) =

(4-2)(3-

1) =

12(1

i)(l

5).

Now let’s look at an application of the

function to the study of rational

numbers mod 1. We say that the fraction m/n is basic if 0 6 m < n. There-

fore q(n) is the number of reduced basic fractions with denominator n; and

the Farey series 3,, contains all the reduced basic fractions with denominator

n or less, as well as the non-basic fraction

The set of all basic fractions with denominator 12, before reduction to

lowest terms, is

Reduction yields

4.9 PHI AND MU 135

and

can

group

these

fractions

their denominators:

What

can

make

this?

Well,

every

divisor

occurs

denomi-

nator, together with all

cp(d)

its

numerators.

The

only denominators that

occur

are

divisors

12.

Thus

dl)

(~(2)

(~(3)

(~(4)

(~(6)

(~(12)

12.

similar thing will obviously happen if

begin with the unreduced

fractions

0 1

rn,

;;;I . . . .

for

any

hence

xv(d)

d\m

(4.54)

said near the beginning

this chapter that

problems

in number

theory

often

require

sums

over

the

divisors

number.

Well,

(4.54)

one

such

sum,

our

claim is vindicated.

(We

will

see

other

examples.)

Now

here’s

curious

fact:

any

function

such

that the

sum

g(m)

x+(d)

d\m

is multiplicative, then

itself is multiplicative. (This result, together with

(4.54)

and the

fact

that

g(m)

= m is obviously multiplicative,

gives

another

reason

why

cp(m)

is multiplicative.)

can

prove

this

curious

fact

in-

duction

The

basis is

easy

because

(1)

Let

m >

and

assume

that

(ml

m2)

(ml

)

(mz)

whenever

and

m=mlmz

andml

Imz,wehave

g(mlm)

f(d)

f(dldz),

d\ml

dl\ml

dz\mz

and

since

all

divisors

are

relatively prime

all

divisors

ml. By the induction

hypothesis,

(dl

d2)

(dl

)

(dr

)

except

possibly

when

and

m2;

hence

obtain

(

f(dl)

f(b))

f(m)f(w)

f(mmz)

\ml dz\m

s(ml)s(mz)

-f(ml)f(m2)

+f(mm2).

But this equals g(mlmr) = g(ml)g(mz),

f(mlm2) = f(ml)f(mr).

136 NUMBER THEORY

Conversely, if f(m) is multiplicative, the corresponding sum-over-divisors

function g(m) =

td,m

f(d) is always multiplicative. In fact, exercise 33 shows

that even more is true. Hence the curious fact is a fact.

The Miibius

finction

F(m),

named after the nineteenth-century mathe-

matician August Mobius who also had a famous band, is defined for all m 3 1

by the equation

p(d)

[m=l].

d\m

(4.55)

This equation is actually a recurrence, since the left-hand side is a sum con-

sisting of p(m) and certain values of p(d) with d < m. For example, if we

plug in m =

2, .

. .

, 12 successively

can compute

the

first twelve values:

n 12

4 5 6

910

cl(n)

1 -1

-1

0 -1 1

-1

0 1

-1 0

Mobius came up with the recurrence formula (4.55) because he noticed

that it corresponds to the following important “inversion principle”:

g(m)

xf(d)

d\m

f(m)

x~(d)g(T)

d\m

(4.56)

According to this principle, the

function gives us a new way to understand

any function f(m) for which we know

Ed,,,,

f(d).

Now is a

good time

The proof of (4.56) uses two tricks (4.7) and (4.9) that we described near

try

WamW

the beginning of this chapter: If g(m) =

td,m

f(d) then

exercise 11.

g(d)

f(k)

k\d

k\m

d\Cm/k)

[m/k=llf(k)

= f(m).

k\m

The other half of (4.56) is proved similarly (see exercise 12).

Relation (4.56) gives us a useful property of the Mobius function, and we

have tabulated the first twelve values; but what is the value of p(m) when

4.9 PHI AND MU 137

m is large? How can we solve the recurrence (4.55)? Well, the function

g(m) = [m =

is obviously multiplicative-after all, it’s zero except when

m = 1. So the Mobius function defined by (4.55) must be multiplicative, by

Depending on bow

what we proved a minute or two ago. Therefore we can figure out what

k(m)

fast you read.

is if we compute p(pk).

When m = pk, (4.55) says that

cl(l)+CL(P)+CL(P2)+...+CL(Pk)

= 0

for all k 3

since the divisors of

are 1, . . . , pk. It follows that

cl(P)

= -1;

p(pk)

= 0 for k > 1.

Therefore by (4.52), we have the general formula

ifm=pjpz...p,;

if m is divisible by some p2.

(4.57)

That’s

If we regard (4.54) as a recurrence for the function

q(m),

we can solve

that recurrence by applying Mobius’s rule (4.56). The resulting solution is

v(m)

Ad):.

d\m

(4.58)

For example,

(~(14 = ~(1)~12+~~(2)~6+~(3)~4+~(4)~3+~(6)~2+~(12)~1

=12-6-4+0+2+0=4.

If m is divisible by

different primes, say

{p,

, . . . , p,},

the sum (4.58) has only

2’ nonzero terms, because the

function is often zero. Thus we can see that

(4.58) checks with formula (4.53), which reads

cp(m)

= m(l

J-) . . .

(I-

J-)

;

if we multiply out the

factors (1

1 /pi), we get precisely the 2’ nonzero

terms of (4.58). The advantage of the Mobius function is that it applies in

many situations besides this one.

For example, let’s try to figure out how many fractions are in the Farey

series 3n. This is the number of reduced fractions in

[O,

whose denominators

do not exceed n, so it is 1 greater than O(n) where we define

Q(x) =

v(k).

l<k<x

(4.59)

138 NUMBER THEORY

(We must add 1 to O(n) because of the final fraction $.) The sum in (4.59)

looks difficult, but we can determine m(x) indirectly by observing that

(4.60)

for all real x 3 0. Why does this identity hold? Well, it’s a bit awesome yet

not really beyond our ken. There are

Lx]11

basic fractions m/n with

0 6 m < n < x, counting both reduced and unreduced fractions; that gives

us the right-hand side. The number of such fractions with gcd(m,n) = d

is @(x/d), because such fractions are

m//n’

with 0 < m’ < n’ 6 x/d after

replacing m by m’d and n by n’d. So the left-hand side counts the same

fractions in a different way, and the identity must be true.

Let’s look more closely at the situation, so that equations (4.59) and

(4.60) become clearer. The definition of m(x) implies that m,(x) =

@(lx]);

but it turns out to be convenient to define m,(x) for arbitrary real values, not

(This extension to

just for integers. At integer values we have the table

real values is a use-

ful trick for many

n 0 12 3 4 5 6 7 8 9 10 11 12

recurrences that

arise in the analysis

v(n)

-112 2 4 2 6 4 6 4 10 4

of algorithms.)

o(n)

0 1 2 4 6 10 12 18 22 28 32 42 46

and we can check (4.60) when x = 12:

@,(12)

D,(6)

+@(4)

f@(3)

+ O(2) +

m,(2)

+6.@,(l)

46+12+6+4+2+2+6

= 78 = t.12.13.

Amazing.

Identity (4.60) can be regarded as an implicit recurrence for

0(x);

for

example, we’ve just seen that we could have used it to calculate

(12) from

certain values of

D(m)

with m < 12. And we can solve such recurrences by

using another beautiful property of the Mobius function:

g(x)

f(x/d)

da1

tr’

(4.61)

This inversion law holds for all functions f such that

tk,da,

If(x/kd)I < 00;

we can prove it as follows. Suppose g(x) =

td3,

f(x/d).

Then

Ad)g(x/d)

Ad)

f(x/kd)

d>l

d>l k>l

f(x/m)

vL(d)[m=kdl

lTt>l

d,kal

4.9 PHI AND MU 139

f(x/m)

p(d)

f(x/m)[m=l]

= f(x).

m>l

d\m

lll>l

The proof in the other direction is essentially the same.

So now we can solve the recurrence (4.60) for

a(x):

D,(x)

;

Ad)

lx/d.lll

x/d1

d>l

(4.62)

This is always a finite sum. For example,

Q(12)

;(12.13-6.7-4.5+0-2.3+2.3

-1~2+0+0+1~2-1~2+0)

78-21-10-3+3-1+1-l

= 46.

In Chapter 9 we’ll see how to use (4.62) to get a good approximation to

m(x);

in fact, we’ll prove that

Q(x)

= -$x2 + O(xlogx).

Therefore the function

O(x)

grows “smoothly”; it averages out the erratic

behavior of

cp(k).

In keeping with the tradition established last chapter, let’s conclude this

chapter with a problem that illustrates much of what we’ve just seen and that

also points ahead to the next chapter. Suppose we have beads of n different

colors; our goal is to count how many different ways there are to string them

into circular necklaces of length m. We can try to “name and conquer” this

problem by calling the number of possible necklaces N (m, n).

For example, with two colors of beads R and B, we can make necklaces

of length 4 in N

(4,2)

= 6 different ways:

f-R\

/R\ fR\

c-R\

/R-\

c-B>

RR RR RB BB BB BB

<R’ <B’

LB’

<R’

LBJ

cBJ

All other ways are equivalent to one of these, because rotations of a necklace

do not change it. However, reflections are considered to be different; in the

case m = 6, for example,

/B-J

f-B>

R R R R

is different from

<BJ

140 NUMBER THEORY

The problem of counting these configurations was first solved by P. A.

Mac-

Mahon in 1892

[212].

There’s no obvious recurrence for N (m, n), but we can count the neck-

laces by breaking them each into linear strings in m ways and considering the

resulting fragments. For example, when m = 4 and n = 2 we get

RRRR

RRRR RRRR RRRR

RRBR RRRB BRRR RBRR

RBBR

RRBB BRRB BBRR

RBRB BRBR RBRB

BRBR

RBBB

BRBB

BBRB BBBR

BBBB BBBB BBBB

BBBB

Each of the

possible patterns appears at least once in this array of

mN(m,n) strings, and some patterns appear more than once. How many

times does a pattern

a~.

. .

a,,-, appear? That’s easy: It’s the number of

cyclic shifts ok . . .

a,-,

a0 . . . ok-1

that produce the same pattern as the orig-

inal a0 . . .

a,-,

For example, BRBR occurs twice, because the four ways to

cut the necklace formed from BRBR produce four cyclic shifts (BRBR, RBRB,

BRBR, RBRB); two of these coincide with BRBR itself. This argument shows

that

mN(m,n) =

[ao...a,_l

=ak...amplaO...ak-l]

q,,...,a,e,ES,

O$k<m

[a0 . .

.a,-,

=ak..

am-lao..

. ak-l] .

O$k<m

ao,...,a,-,ES,

Here S, is a set of n different colors.

Let’s see how many patterns satisfy a0 . . .

a,-1

= ok. . .

a,-,

a0 . . .

ok-l,

when k is given. For example, if m = 12 and k = 8, we want to count the

number of solutions to

This means

a4;

al =

= as;

= alo =

o6;

and

all

a7.

So the values of ao,

al,

a2, and

can be chosen in n4 ways, and the remaining

a’s depend on them. Does this look familiar? In general, the solution to

%+k)modm

for 0 < j < m

makes US equate

with

o(i+kr)

modm

for 1 = 1, 2, . .

and we know that

the multiples of k modulo m are

(0,

d, 2d,. . . , m

d}, where d = gcd(k, m).

Therefore the general solution is to choose

ao,

. . . , o&l independently and

then to set

oj+d

for d < j < m. There are nd solutions.

4.9 PHI AND MU 141

We have just proved that

mN(m,n) =

ngcdCkVm)

O<k<m

This sum can be simplified, since it includes only terms nd where d\m. Sub-

stituting d = gcd(k, m) yields

N(m,n) =

[d=gcd(k,m)]

d\m

O<k<m

nd~

[k/d.l m/d]

d\m

O<k<m

[kIm/d].

d\m

O<k<m/d

(We are allowed to replace k/d by k because k must be a multiple of d.)

Finally, we have

&‘,,,,,/d [klm/d]

cp(m/d)

by definition, so we obtain

MacMahon’s

formula:

N(m,n) =

d,mndg(T)

ixdd)nm/d

d\m

(4.63)

When m = 4 and n = 2, for example, the number of necklaces is

.24

.22

+ 2.2’) = 6, just as we suspected.

It’s not immediately obvious that the value N(m, n) defined by

Mac-

Mahon’s sum is an integer! Let’s try to prove directly that

cp(d)nm’d

d\m

(mod m),

(4.64)

without using the clue that this is related to necklaces. In the special case

that m is prime, this congruence reduces to n” + (p

1)n = 0 (mod p); that

is, it reduces to np = n. We’ve seen in (4.48) that this congruence is an

alternative form of Fermat’s theorem. Therefore (4.64) holds when m = p;

we can regard it as a generalization of Fermat’s theorem to the case when the

modulus is not prime. (Euler’s generalization (4.50) is different.)

We’ve proved (4.64) for all prime moduli, so let’s look at the smallest

case left, m = 4. We must prove that

n4+n2+2n

= 0

(mod 4) .

The proof is easy if we consider even and odd cases separately. If n is even,

all three terms on the left are congruent to 0 modulo 4, so their sum is too. If

142 NUMBER THEORY

n is odd,

and

are each congruent to 1, and 2n is congruent to 2; hence

the left side is congruent to I + 1

and thus to 0 modulo 4, and we’re done.

Next, let’s be a bit daring and try m = 12. This value of m ought to

be interesting because it has lots of factors, including the square of a prime,

yet it is fairly small. (Also there’s a good chance we’ll be able to generalize a

proof for 12 to a proof for general m.) The congruence we must prove is

n”+n6+2n4+2n3+2n2+4n

E 0 (mod 12).

Now what? By (4.42) this congruence holds if and only if it also holds mod-

ulo 3 and modulo 4. So let’s prove that it holds modulo 3. Our congru-

ence (4.64) holds for primes, so we have

+ 2n = 0 (mod 3). Careful

scrutiny reveals that we can use this fact to group terms of the larger sum:

n’2+n6+2n4+2n3+2n2+4n

(n12

+2n4) +

In6

+2n2)

+2(n3

+2n)

0+0+2*0

5 0

(mod

3).

So it works modulo 3.

We’re half done. To prove congruence modulo 4 we use the same trick.

We’ve proved that

+n2

+2n

= 0 (mod 4), so we use this pattern to group:

n”+n6+2n4+2n3+2n2+4n

(n12

+ 2n3) + 2(n4 +

+ 2n)

0+2.0

E 0 (mod 4).

QED for the case m = 12.

QED: Quite

Easily

So far we’ve proved our congruence for prime m, for m = 4, and for m =

Done.

12. Now let’s try to prove it for prime powers. For concreteness we may

suppose that m =

for some prime p. Then the left side of (4.64) is

np3

cp(p)nP2

+ q(p2)nP + cp(p3)n

np3

+ (p

)np2

(p2

p)nP +

(p3

p2)n

(np3

npz) +

p(np2

nP) + p2(nP

-n)

+p3n.

We can show that this is congruent to 0 modulo

if we can prove that

n’J3

nP2

is divisible by

p3,

that

nP2

is divisible by

p2,

and that n”

is divisible by p, because the whole thing will then be divisible by

p3.

By the

alternative form of Fermat’s theorem we have

E n (mod p), so p divides

n; hence there is an integer q such that

nfpq

4.9 PHI AND MU 143

Now we raise both sides to the pth power, expand the right side according to

the binomial theorem (which we’ll meet in Chapter 5), and regroup, giving

TIP2

+ pq)p =

+ (pq)‘nPm’

+ (pq)2nPP2

0 0

+ p2Q

for some other integer Q. We’re able to pull out a factor of

here because

= p in the second term, and because a factor of (pq)’ appears in all the

terms that follow. So we find that

divides

npz

np.

Again we raise both sides to the pth power, expand, and regroup, to get

np3

(nP

+ P~Q)~

nP2 +

(p2Q)‘nP’Pp’l

(p2Q)2nP’P-2’

+ . .

np2

+ p3Q

for yet another integer Q. So

divides

nP3-

np’. This finishes the proof for

m = p3, because we’ve shown that

divides the left-hand side of (4.64).

Moreover we can prove by induction that

n~k

n~km’ +

pkD

for some final integer

(final because we’re running out of fonts); hence

nPk

nPk-’

(mod

~~1,

for

> 0.

(4.65)

Thus the left side of (4.64), which is

Pk-nPkm’)

p(nPkm’-nPkmZ)

+ . . .

pkpl(nP-,)

pkn,

is divisible by

and so is congruent to 0 modulo

pk.

We’re almost there. Now that we’ve proved (4.64) for prime powers, all

that remains is to prove it when m = m’ m2, where m’

ml,

assuming that

the congruence is true for m’ and m2. Our examination of the case m = 12,

which factored into instances of m = 3 and m = 4, encourages us to think

that this approach will work.

We know that the

function is multiplicative, so we can write

q(d)nm’d

(P(d’d2)nm1mz’d1d2

d\m

\ml>

dr\mz

oldl)(

di\ml

dz\mz

144 NUMBER THEORY

But the inner sum is congruent to 0 modulo mz, because we’ve assumed that

(4.64) holds for ml; so the entire sum is congruent to 0 modulo

m2.

By a

symmetric argument, we find that the entire sum is congruent to 0 modulo

as well. Thus by (4.42) it’s ‘congruent to 0 modulo m. QED.

Exercises

Warmups

What is the smallest positive integer that has exactly k divisors, for

l<k$6?

Prove that gcd( m, n) . lcm( m, n) =

m.n,

and use this identity to express

lcm(m,n) in terms of

lc.m(n

mod m, m), when n mod m # 0. Hint: Use

(4.121,

(4.14))

and

(4.15).

Let

71(x)

be the number of primes not exceeding x. Prove or disprove:

n(x)

X(X

1) =

is prime]

4 What would happen if the Stern-Brocot construction started with the

five fractions

(p,

e) instead of with (f,

$)?

Find simple formulas for

and

Rk,

when L and R are the 2 x 2 matrices

of (4.33).

What does ‘a = b (mod 0)’ mean?

Ten people numbered

to 10 are lined up in a circle as in the

Josephus

problem, and every mth person is executed. (The value of m may be

much larger than 10.) Prove that the first three people to go cannot be

10, k, and

1 (in this order), for any k.

8 The residue number system (x mod 3, x mod 5) considered in the text has

the curious property that 13 corresponds to

(1,3),

which looks almost the

same. Explain how to find all instances of such a coincidence, without

calculating all fifteen pairs of residues. In other words, find all solutions

to the congruences

lOx+y G x (mod3), lOx+y E y (mod5).

Hint: Use the facts that lOu+6v = u (mod 3) and lOu+6v = v (mod 5).

9 Show that (3”

1)/2

is odd and composite. Hint: What is

3”

mod

10 Compute

(~(999).

4 EXERCISES 145

Find a function o(n) with the property that

g(n) =

f(k)

f(n) =

o(k)g(n-k).

O<k<n O<k<n

(This is analogous to the Mobius function; see (4.56).)

12 Simplify the formula

xd,,,,

tkjd

F(k) g(d/k).

13 A positive integer n is called squarefree if it is not divisible by m2 for

any m > 1. Find a necessary and sufficient condition that n is squarefree,

in terms of the prime-exponent representation (4.11) of n;

b in terms of u(n).

Basics

14 Prove or disprove:

gcd(km, kn) = kgcd(m,n) ;

b lcm(km, kn) =

klcm(m,n)

15 Does every prime occur as a factor of some Euclid number e,?

16 What is the sum of the reciprocals of the first n Euclid numbers?

17 Let

be the “Fermat number”

22”

+ 1. Prove that

if m < n.

18 Show that if 2” + 1 is prime then n is a power of 2.

For every positive integer n there’s a prime p such that n < p 6 2n. (This

is essentially “Bertrand’s postulate,” which Joseph Bertrand verified for

n < 3000000 in 1845 and Chebyshev proved for all n in 1850.) Use

Bertrand’s postulate to prove that there’s a constant b

1.25 such that

the numbers

129,

1227,

[2q

. . .

are all prime.

20 Let

be the nth prime number. Find a constant K such that

[(10n2K)

mod

10n]

= P,.

21 Prove the following identities when n is a positive integer:

Hint: This is a trick question and the answer is pretty easy.

146 NUMBER THEORY

22 The number 1111111111111111111 is prime. Prove that, in any radix b,

Is this a test for

(11 . . . 1

can be prime only if the number of 1

‘s

is prime.

strabismus?

State a recurrence for p(k), the ruler function in the text’s discussion of

ez(n!).

Show that there’s a connection between p(k) and the disk that’s

moved at step k when an n-disk Tower of Hanoi is being transferred in

1 moves, for 1

24 Express e,(n!) in terms of y,,(n), the sum of the digits in the radix p

Look, ma,

representation of n, thereby generaliZing

(4.24).

sideways addition.

We say that m esactly divides n, written m\\n, if m\n and m J- n/m.

For example, in the text’s discussion of factorial factors,

p”P(“!)\\n!.

Prove or disprove the following:

k\\n and m\\n

km\\n, if k

b For all m,n > 0, either gcd(m, n)\\m or gcd(m, n)\\n.

Consider the sequence

of all nonnegative reduced fractions m/n such

that mn 6 N For example,

cJIO

11111111

z s

1'10'9'8'7'b'5'4'3'5'2'3'1'2'1'2'1'2'1'1'~'1'1'1'1'

Is it true that m’n

mn’ = 1 whenever m/n immediately precedes

m//n’ in

$Y!N?

27 Give a simple rule for

c:omparing

rational numbers based on their repre-

sentations as L’s and R’s in the Stern-Brocot number system.

28 The Stern-Brocot representation of

R3L7R’5LR29i’LRLR2LR3LR14L2R,.

;

use it to

find

all the simplest rational approximations to

whose denom-

inators are less than 50. Is

one of them?

29 The text describes a correspondence between binary real numbers

(.blb2b3..

in [0, 1) and Stern-Brocot real numbers

B2B3

. . . in

[O,

00).

If x corresponds to

and

# 0, what number corresponds to

l--x?

30 Prove the following statement (the Chinese Remainder Theorem): Let

ml,

. . . .

m, be integers with

for 1 6 j < k < r; let m =

. . .

m,; and let

al,

. . . .

arr

A be integers. Then there is exactly one

integer a such that

a=ak(modmk)fOrl<k<r

and

A<a<A+m.

A number in decimal notation is divisible by 3 if and only if the sum of

its digits is divisible by 3. Prove this well-known rule, and generalize it.

4 EXERCISES 147

Why is “Euler”

32 Prove Euler’s theorem (4.50) by generalizing the proof of (4.47).

pronounced “Oiler”

when “Euclid” is

33 Show that if f(m) and g(m) are multiplicative functions, then so is

“Yooklid”?

h(m) =

tdim

f(d)

g(m/d).

Prove that (4.56) is a special case of (4.61).

Homework exercises

Let I(m,n) be a function that satisfies the relation

I(m,n)m+ I(n,m)n = gcd(m,n),

when m and n are nonnegative integers with m # n. Thus, I(

n) = m’

and

I(n,

m) = n’ in (4.5); the value of

I(m,

n) is an inverse of m with

respect to n. Find a recurrence that defines I(m,n).

36 Consider the set

Z(m)

= {m +

n&?

1 integer m,n}. The number

m +

is called a unit if

On2

since it has an inverse

(that is, since

(m+nm).+(m-n&?)

= 1). For example,

3+mis

a unit, and so is 19

6m.

Pairs of cancelling units can be inserted into

any factorization, so we ignore them. Nonunit numbers of

Z(m

)

are

called prime if they cannot be written as a product of two nonunits. Show

that2,3,and4fnareprimesofZ(fl).

Hint:

If2=(k+L&?)x

(m +

n&?

) then 4 =

(kz

012)

(

On’).

Furthermore, the square

of any integer mod 10 is 0, 1, 4, 5, 6, or 9.

37 Prove (4.17). Hint: Show that

i = (e,_l

i)’

and consider

2-nlog(e,

t).

Prove that if a

b and a > b then

gcd(am _

bm,

_ bn) =

agcd(m>n)

_ bdm>ni

O$m<n.

(All variables are integers.) Hint: Use Euclid’s algorithm.

39 Let S(m) be the smallest positive integer n for which there exists an

increasing sequence of integers

m =

< a2 < ... < at = n

such that al

al..

. at is a perfect square. (If m is a perfect square, we

can let t = 1 and n =

m.)

For example, S(2) = 6 because the best such

sequence is 2.3.6. We have

n 12345

6 7 8 9 10

S(n)

1 6 8 4 10

12 14 15 9 18

22 20

Prove that S(m) #

(m’) whenever 0 < m < m’.

148 NUMBER THEORY

40 If the radix p representation of n is (a,,, . . . al ao)v, prove that

epCn!)

(-l)“P(n!‘a,!.

. . a,! ao! (mod p)

(The left side is simply n! with all p factors removed. When n = p this

reduces to Wilson’s theorem.)

Show that if p mod. 4 = 3, there is no integer n such that p divides

n* + 1. Hint: Use :Fermat’s theorem.

b But show that if p mod 4 =

there is such an integer. Hint: Write

I)!

(II,=,

‘p~‘i’2 k(p

k)) and think about Wilson’s theorem.

Consider two fractions

m/n

and m//n’ in lowest terms. Prove that when

the sum

m/n+m’/n’

is reduced to lowest terms, the denominator will be

nn’ if and only if n

n’. (In other words, (mn’+m’n)/nn’ will already

be in lowest terms if and only if n and n’ have no common factor.)

43 There are

nodes at level k of the Stern-Brocot tree, corresponding to

the matrices Lk Lkp’ R

..I

Rk.

Show that this sequence can be obtained

by starting with Lk and’then multiplying successively by

-1

2p(n)

+ 1

for 1 6 n < 2k, where p(n) is the ruler function.

44 Prove that a baseball player whose batting average is .316 must have

batted at least 19 times. (If he has m hits in n times at bat, then

m/n

[.3155, .3165).)

45 The number 9376 has the peculiar self-reproducing property that

9376*

= 87909376

How many 4-digit numbers x satisfy the equation

mod 10000 = x?

How many n-digit numbers x satisfy the equation

mod

10n

= x?

Prove that if nj =

and nk = 1 (mod m), then nscd(jtk) = 1.

b Show that 2”

1 (mod n), if n > 1. Hint: Consider the least prime

factor of n.

Show that if nmp’ E 1 (mod m) and if n(“-‘)/p $ 1 (mod m) for all

primes such that p\(m

l), then m is prime. Hint: Show that if this

condition holds, the numbers nk mod m are distinct, for 1

k < m.

Generalize Wilson’s theorem (4.49) by ascertaining the value of the ex-

pression u-I1 <n<m, nlm

n)modm,whenm>l.

Wilson’s theorem:

“Martha, that boy

is a menace.”

Radio announcer:

‘I

. . .

pitcher Mark

LeChiffre

hits a

two-run single!

Mark was batting

only .080, so he gets

his

second

hit of

the year.

”

Anything wrong?

The proof that large

numbers are prime

is very easy: Let

be a

large prime

number; then x is

prime, QED.

4 EXERCISES 149

49 Let R(N) be the number of pairs of integers (m, n) such that 0 6 m < N,

O<n<N,andmIn.

Express R(N) in terms of the

function.

Prove that R(N) = EdaN LN/dJ’p(d).

50 Let m be a positive integer and let

e2nilm

= cos(2n/m) +isin(27r/m).

What

are the roots

of disunity?

We say that w is an mth root of unity, since

eZni

= 1. In fact,

each of the m complex numbers

w”,

w’,

. . , w”-’ is an mth root of

unity, because

(wk)“’

eZnki

= 1; therefore

is a factor of the

polynomial

1, for 0 < k < m. Since these factors are distinct, the

complete factorization of

1 over the complex numbers must be

-1 =

(Z-Wk).

O<k<m

a Let

Y,(z)

Oik<m,klm(~

wk).

(This polynomial of degree

q(m) is called the cyclotomic polynomial of order

m.)

Prove that

-1 = r-p&(Z).

d\m

Prove that

Ym(z)

= nd,m(~d

l)k(m/d).

Exam problems

Prove Fermat’s theorem (4.48) by expanding

1 +

1)P

via the

multinomial theorem.

Let n and x be positive integers such that

has no divisors 6 n (except

l),

and let p be a prime number. Prove that at least

Ln/p]

of the numbers

{X-l,X2-1,...,Xn~'

}

are multiples of

Find all positive integers n such that n \ [(n

l)!/(n +

l)].

54 Determine the value of

lOOO!

mod

1O25o

by hand calculation.

55 Let

be the product of the first n factorials,

ni=,

k!. Prove that

P2,/PP,

is an integer, for all positive integers n.

56 Show that

2np1

n-1

I-I

pin(k,

Zn-k)

I-n

2k+

ZnpZk-1

k=l k=l

is a power of 2.

150 NUMBER THEORY

57 Let S(m,n) be the set of all integers k such that

mmodk+nmodk 3 k.

For example, S(7,9) =

{2,4,5,8,10,11,12,13,14,15,16}.

Prove that

q(k) = m.n

kESlm,n)

Hint: Prove first that

x,6msn

,&,,,

v(d) =

IL>,

v(d)

ln/dJ.

Then

consider

L(m

+ n)/d]

[m/d]

Ln/dJ.

58 Let f(m) =

Ed,,,,

Fi:nd

a necessary and sufficient condition that f(m)

is a power of 2.

Bonus problems

59 Prove that if x1, . . . ,

are positive integers with 1 /x1

. . + 1

/x,

then

max(xl,.

. . ,x,) < e,. Hint: Prove the following stronger result by

induction: “If 1 /x1

. . + 1

/x,

l/o1

where x1, . . . ,

are positive

integers and

is a rational number 3

max(xl

, . . , xn), then a+ 1 <

e,+l

and

(a + 1) <

. . .

e,e,+l

.”

(The proof is nontrivial.)

60 Prove that there’s a constant P such that (4.18) gives only primes. You

may use the following

(Ihighly

nontrivial) fact: There is a prime between

p and p + cp’, for some constant c and all sufficiently large p, where

g=losl.

1920

61 Prove that if m/n,

m’/n’,

and

m/‘/n”

are consecutive elements of 3~,

then

m” =

[(n+N)/n’]m’-m,

n” =

[(n+N)/n’jn’-n.

(This recurrence allows us to compute the elements of

in order, start-

ing with

and ft.)

62 What binary number corresponds to e, in the binary

Stern-Brocot

correspondence? (Express your answer as an infinite sum; you need not

evaluate it in closed form.)

63 Show that if Fermat’s Last Theorem (4.46) is false, the least n for which

it fails is prime. (You may assume that the result holds when n = 4.)

Furthermore, if

and a

b, show that there exists an integer

m such that

a+b

mp,

if p$c;

pPV1 mP

if p\c.

Thus c must be really huge. Hint: Let x = a + b, and note that

gcd(x,

(ap

+ (x

a)p)/x)

= gcd(x,paP-‘).

4 EXERCISES 151

64 The Peirce sequence

3’~

of order N is an infinite string of fractions

separated by ‘<’ or ‘=’

signs, containing all the nonnegative fractions

m/n with m > 0 and n 6 N (including fractions that are not reduced).

It is defined recursively by starting with

For N > 1, we form ?$,+I by inserting two symbols just before the kNth

symbol of ?N, for all k > 0. The two inserted symbols are

k-l

N+l

’

if kN is odd;

k-l

yN,kN

N+l’

if kN is even.

Here ?N,j denotes the jth symbol of

Y’

which will be either ‘<’ or ‘=’

when j is even; it will be a fraction when j is odd. For example,

Ip2

~=~<t<f=f<I<4=f<5<4=~~~~~=~~~~~=~~...;

4=~=P<~<t<3<~=~=t<~<~<~~~=~=~~~~~~...~

4=~=Q=q<1,1,2=L,2,3,~=~=~=~~~~~~~=,..;

4 3 4 2 3 4 2 4 3

Ip5

~=~=P=Q=q<l<l<l<r<l=1,1,2,3,1,2=4=....;

543542534524

Ip6

q,~,~,g,Q=~<l,l,l,l,l,Z,1=3=L,3,4=....

65463546256

(Equal elements occur in a slightly peculiar order.) Prove that the ‘<’

and ‘=’ signs defined by the rules above correctly describe the relations

between adjacent fractions in the Peirce sequence.

Research problems

65 Are the Euclid numbers

all squarefree?

66 Are the Mersenne numbers

1 all squarefree?

67 Prove or disprove that maxl<j<kbn ok/gCd(oj, ok) 3 n, for all sequences

of integers 0 < al <

...

< a,.

68 Is there a constant Q such that [Q’“] is prime for all n 3 O?

69 Let

denote the nth prime. Prove or disprove that P,+r

O(logP,)?

70 Does es(n!) = ez(n!)/2 for infinitely many n?

71 Prove or disprove: If k # 1 there exists n > 1 such that 2”

k (mod n).

Are there infinitely many such n?

72 Prove or disprove: For all integers a, there exist infinitely many n such

that

cp(n)\(n

+ a).

152 NUMBER THEORY

If the 0(n) + 1 terms of the Farey series

were fairly evenly distributed, we would expect 3n(k)

k/@(n). There-

fore the sum D(n) =

~~~‘[3~(k)

k/O(n)1 measures the “deviation

3,,

from uniformity!’ Is it true that D(n) = 0

(n1/2+E)

for all

> O?

74 Approximately how many distinct values are there in the set {O! mod p,

l!modp,...,(p-l)!modp},asp+oo?

Binomial Coefficients

Lucky us!

Otherwise known

as combinations of

n things, k at a

time.

LET’S TAKE A BREATHER. The previous chapters have seen some heavy

going, with sums involving floor, ceiling, mod, phi, and mu functions. Now

we’re going to study binomial coefficients, which turn out to be (a) more

important in applications, and (b) easier to manipulate, than all those other

quantities.

5.1

BASIC IDENTITIES

The symbol (t) is a binomial coefficient, so called because of an im-

portant property we look at later this section, the binomial theorem. But we

read the symbol “n choose

k!’

This incantation arises from its combinatorial

interpretation-it is the number of ways to choose a k-element subset from

an n-element set. For example, from the set {1,2,3,4} we can choose two

elements in six ways,

(“2)

= 6.

To express the number (c) in more familiar terms it’s easiest to first

determine the number of k-element sequences, rather than subsets, chosen

from an n-element set; for sequences, the order of the elements counts. We

use the same argument we used in Chapter 4 to show that n! is the number

of permutations of n objects. There are n choices for the first element of the

sequence; for each, there are n-l choices for the second; and so on, until there

are

n-k+1

choices for the kth. This gives n(n-1). . .

(n-k+l)

= nk choices

in all. And since each k-element subset has exactly k! different orderings, this

number of sequences counts each subset exactly k! times. To get our answer,

we simply divide by k!:

n(n-l)...(n-k+l)

k(k-l)...(l)

’

153

154 BINOMIAL COEFFICIENTS

For example,

4 2

=-=

4.3

2.1

this agrees with our previous enumeration.

We call n the upper index and k the lower index, The indices are

restricted to be nonnegative integers by the combinatorial interpretation, be-

cause sets don’t have negative or fractional numbers of elements. But the

binomial coefficient has many uses besides its combinatorial interpretation,

so we will remove some of the restrictions. It’s most useful, it turns out,

to allow an arbitrary real (or even complex) number to appear in the upper

index, and to allow an arbitrary integer in the lower. Our formal definition

therefore takes the following form:

r(r-l)...(r-kkl)

r-k

k(k-l)...(l) = k!’

integer k 3 0;

(5.1)

integer k < 0.

This definition has several noteworthy features. First, the upper index is

called

not n; the letter r emphasizes the fact that binomial coefficients make

sense when any real number appears in this position. For instance, we have

(,')

= (-l)(-2)(-3)/(3.2.1)= -1.

ere’s

no combinatorial interpretation

here, but r = -1 turns out to be an important special case. A noninteger

index like r =

-l/2

also turns out to be useful.

Second, we can view

(;>I

as a kth-degree polynomial in

We’ll see that

this viewpoint is often helpful.

Third, we haven’t defined binomial coefficients for noninteger lower in-

dices. A reasonable definition can be given, but actual applications are rare,

so we will defer this generalization to later in the chapter.

Final note: We’ve listed the restrictions ‘integer k 3 0’ and ‘integer

k < 0’ at the right of the definition. Such restrictions will be listed in all

the identities we will study, so that the range of applicability will be clear.

In general the fewer

restricti.ons

the better, because an unrestricted identity

is most useful; still, any restrictions that apply are an important part of

the identity. When we manipulate binomial coefficients, it’s easier to ignore

difficult-to-remember restrictions temporarily and to check later that nothing

has been violated. But the check needs to be made.

For example, almost every time we encounter

(“,)

it equals 1, so we can

get lulled into thinking that it’s always 1. But a careful look at definition (5.1)

tells us that

(E)

is 1 only when n

0 (assuming that n is an integer); when

n < 0 we have

(“,)

= 0. Traps like this can (and will) make life adventuresome.

5.1 BASIC IDENTITIES 155

Binomial coefficients

were well known

Asia, many

cen-

turies before Pascal

was born

1741,

but

he bad no way to

know that.

Italy it’s called

Tartaglia’s

triangle.

Before getting to the identities that we will use to tame binomial coeffi-

cients, let’s take a peek at some small values. The numbers in Table 155 form

the beginning of Pascal’s triangle, named after Blaise Pascal (1623-1662)

Table 155 Pascal’s triangle.

12 1

13 3 1

14 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

45 120 210 252 210 120 45 10 1

because he wrote an influential treatise about them

[227].

The empty entries

in this table are actually O’s, because of a zero in the numerator of (5.1); for

example, (l) = (

1.0)/(2.1)

= 0. These entries have been left blank simply to

help emphasize the rest of the table.

It’s worthwhile to memorize formulas for the first three columns,

=I,

(;)=?.,

(;)2g;

these hold for arbitrary reals. (Recall that (“T’) =

in(n

+ 1) is the formula

we derived for triangular numbers in Chapter 1; triangular numbers are con-

spicuously present in the (;) column of Table 155.) It’s also a good idea to

memorize the first five rows or so of Pascal’s triangle, so that when the pat-

tern

4, 6, 4, 1 appears in some problem we will have a clue that binomial

coefficients probably lurk nearby.

The numbers in Pascal’s triangle satisfy, practically speaking, infinitely

many identities, so it’s not too surprising that we can find some surprising

relationships by looking closely. For example, there’s a curious “hexagon

property,” illustrated by the six numbers 56, 28, 36, 120, 210, 126 that sur-

round 84 in the lower right portion of Table 155. Both ways of multiplying

alternate numbers from this hexagon give the same product: 56.36.210 =

28.120.126 = 423360. The same thing holds if we extract such a hexagon

from any other part of Pascal’s triangle.

156 BINOMIAL COEFFICIENTS

And now the identities,. Our goal in this section will be to learn a few

“C’est

une

chose

simple rules by which we can solve the vast majority of practical problems

estrange combien

involving binomial coefficients.

il est fertile en

Definition (5.1) can be recast in terms of factorials in the common case

proprietez.

”

-B.

Pascal /227/

that the upper index

is an integer, n, that’s greater than or equal to the

lower index k:

k = k!(n-k)!’

integers n 3 k

(5.3)

To get this formula, we just multiply the numerator and denominator of (5.1)

by (n

k)!. It’s occasionally useful to expand a binomial coefficient into this

factorial form (for example, when proving the hexagon property). And we

often want to go the other way, changing factorials into binomials.

The factorial representation hints at a symmetry in Pascal’s triangle:

Each row reads the same left-to-right as right-to-left. The identity reflecting

this-called the symmetry identity-is obtained by changing k to n

(5.4)

This formula makes combinatorial sense, because by specifying the k chosen

things out of n we’re in effect specifying the n

k unchosen things.

The restriction that n and k be integers in identity (5.4) is obvious, since

each lower index must be an integer. But why can’t n be negative? Suppose,

for example, that n = -1. Is

(‘)

’

(-ilk)

a valid equation? No. For instance, when k = 0 we get 1 on the left and 0 on

the right. In fact, for any integer k 3 0 the left side is

c-1

I(-2).

.1:-k)

(-,

which is either 1 or -1; but the right side is 0, because the lower index is

negative. And for negative k the left side is 0 but the right side is

(-I)-’

which is either 1 or -1. So the equation ‘(-,‘) =

((;!,)I

is always false!

The symmetry identity fails for all other negative integers n, too. But

unfortunately it’s all too easy to forget this restriction, since the expression

in the upper index is sometimes negative only for obscure (but legal) values

5.1 BASIC IDENTITIES 157

I just hope

don’t

of its variables. Everyone who’s manipulated binomial coefficients much has

fall into this trap

during the

midterm.

fallen into this trap at least three times.

But the symmetry identity does have a big redeeming feature: It works

for all values of k, even when k < 0 or k > n. (Because both sides are zero in

such cases.) Otherwise 0 < k 6 n, and symmetry follows immediately from

(5.3):

k = k!(n-k)!

Our next important

coefficients:

(n-(n--l\!

(n-k)! =

identity lets us move things in and out of binomial

I,(:::))

integer k # 0.

(5.5)

The restriction on k prevents us from dividing by 0 here. We call (5.5)

an absorption identity, because we often use it to absorb a variable into a

binomial coefficient when that variable is a nuisance outside. The equation

follows from definition

(5.1),

because

r(r-

and k! = k(k- l)! when

0; both sides are zero when k < 0.

If we multiply both sides of (5.5) by k, we get an absorption identity that

works even when k = 0:

k(l[)

r(;-i)

, integer k.

This one also has a companion that keeps the lower index intact:

(r-k)(I)

r(‘i’),

integer k.

(5.6)

(5.7)

We can derive (5.7) by sandwiching an application of (5.6) between two ap-

plications of symmetry:

(r-k)(;)

(r-kl(rlk)

(by

symmetry)

r(,.Ti!

(by

(54)

(by symmetry)

But wait a minute. We’ve claimed that the identity holds for all real

yet the derivation we just gave holds only when r is a positive integer. (The

upper index r

1 must be a nonnegative integer if we’re to use the symmetry

158 BINOMIAL COEFFICIENTS

property (5.4) with impunity.) Have we been cheating? No. It’s true that

(We/l,

not here

the derivation is valid only for positive integers r; but we can claim that the

anyway)

identity holds for all values of r, because both sides of (5.7) are polynomials

in r of degree k + 1. A nonzero polynomial of degree d or less can have at

most d distinct zeros; therefore the difference of two such polynomials, which

also has degree d or less, cannot be zero at more than d points unless it is

identically zero. In other words, if two polynomials of degree d or less agree

at more than d points,

the,y

must agree everywhere. We have shown that

(r-k)(;) =

&‘)

w enever

is a positive integer; so these two polynomials

agree at infinitely many points, and they must be identically equal.

The proof technique in the previous paragraph, which we will call the

polynomial argument, is useful for extending many identities from integers

to reals; we’ll see it again and again. Some equations, like the symmetry

identity (5.4), are not identities between polynomials, so we can’t always use

this method. But many identities do have the necessary form.

For example, here’s another polynomial identity, perhaps the most im-

portant binomial identity of all, known as the addition formula:

(‘*‘)

+ (;-I:)

integer k.

(5.8)

When

is a positive integer, the addition formula tells us that every number

in Pascal’s triangle is the sum of two numbers in the previous row, one directly

above it and the other just to the left. And the formula applies also when

is negative, real, or complex; the only restriction is that k be an integer, so

that the binomial coefficients are defined.

One way to prove the addition formula is to assume that

is a positive

integer and to use the combinatorial interpretation. Recall that (I) is the

number of possible k-element subsets chosen from an r-element set. If we

have a set of

eggs that includes exactly one bad egg, there are (i) ways to

select k of the eggs. Exactly

(‘i’)

of these selections involve nothing but good

eggs; and (,“\) of them contain the bad egg, because such selections have k-l

of the

-- 1 good eggs. Adding these two numbers together gives (5.8). This

derivation assumes that

is a positive integer, and that k 3 0. But both sides

of the identity are zero when k < 0, and the polynomial argument establishes

(5.8) in all remaining cases.

We can also derive (5.8) by adding together the two absorption identities

(5.7) and (5.6):

(r-k)(;)

+k(l)

r(‘i’)

+r(;-:);

the left side is r(i), and we can divide through by

This derivation is valid

for everything but r = 0, and it’s easy to check that remaining case.

5.1 BASIC IDENTITIES 159

Those of us who tend not to discover such slick proofs, or who are oth-

erwise into tedium, might prefer to derive (5.8) by a straightforward manip-

ulation of the definition. If k > 0,

(r- l)k

(r-

l)k-’

(‘*‘)+(;I:)

k!+

(k- l)!

(T-l)lf=l(r-k)

(r-l)k-‘k

(r-l)Er

k! k!

’

Again, the cases for k < 0 are easy to handle.

We’ve just seen three rather different proofs of the addition formula. This

is not surprising; binomial coefficients have many useful properties, several of

which are bound to lead to proofs of an identity at hand.

The addition formula is essentially a recurrence for the numbers of Pas-

cal’s triangle, so we’ll see that it is especially useful for proving other identities

by induction. We can also get a new identity immediately by unfolding the

recurrence. For example,

(Z)

(;)

(Z)

(D+(i)+(f)

(;)+(;)+(;)+(i)

(I)++++,

Since

(!,)

= 0, that term disappears and we can stop. This method yields

the general formula

,5-,(‘:“)

(a)

(‘7’)

+...+

(“n”)

(r’:“))

integer n.

(5.9)

Notice that we don’t need the lower limit k 3 0 on the index of summation,

because the terms with k < 0 are zero.

This formula expresses one binomial coefficient as the sum of others whose

upper and lower indices stay the same distance apart. We found it by repeat-

edly expanding the binomial coefficient with the smallest lower index: first

160 BINOMIAL COEFFICIENTS

(3, then (i), then (i), then (i). What happens if we unfold the other way,

repeatedly expanding the one with largest lower index? We get

(;)

(l)

(Z)

(i)+(i)+(l)

(i)+(:)+(z)+(:)

(;)+(l)+(:)+(;)+(;)

(i)+(;)+(;)+(;)+(;)+(;)*

Now

(3”)

is zero (so are (i)

a.nd

(i) , but these make the identity nicer), and

we can spot the general pattern:

(&(L)

(e)

(A)

+...+(z)

n+l

(

)

m+l

1 ,

integers

n 3 0.

(5.10)

This identity, which we call summation on the upper index, expresses a

binomial coefficient as the sum of others whose lower indices are constant. In

this case the sum needs the lower limit k 3 0, because the terms with k < 0

aren’t zero. Also, m and n can’t in general be negative.

Identity (5.10) has an interesting combinatorial interpretation. If we want

to choose m + 1 tickets

from1

a set of n + 1 tickets numbered 0 through n,

there are (k) ways to do this when the largest ticket selected is number k.

We can prove both (5.9) and (5.10) by induction using the addition

formula, but we can also prove them from each other. For example, let’s

prove

(5.9)

from (5.10); our proof will illustrate some common binomial co-

efficient manipulations. Our general plan will be to massage the left side

(‘+kk)

(5.9)

so that it looks like the left side

(L) of (5.10); then we’ll

invoke that identity, replacing the sum by a single binomial coefficient; finally

we’ll transform that coefficient into the right side of (5.9).

We can assume for convenience that r and n are nonnegative integers;

the general case of

(5.9)

follows from this special case, by the polynomial

argument. Let’s write m instead of

so that this variable looks more like

a nonnegative integer. The

:plan

can now be carried out systematically as

5.1 BASIC IDENTITIES 161

Let’s look at this derivation blow by blow. The key step is in the second line,

where we apply the symmetry law (5.4) to replace (“,‘“) by (“‘,‘“). We’re

allowed to do this only when m + k 3 0, so our first step restricts the range

of k by discarding the terms with k < -m. (This is legal because those terms

are zero.) Now we’re almost ready to apply (5.10); the third line sets this up,

replacing k by k

m and tidying up the range of summation. This step, like

the first, merely plays around with t-notation. Now k appears by itself in

the upper index and the limits of summation are in the proper form, so the

fourth line applies (5.10). One more use of symmetry finishes the job.

Certain sums that we did in Chapters 1 and 2 were actually special cases

(5.10),

or disguised versions of this identity. For example, the case m = 1

gives the sum of the nonnegative integers up through n:

(;)

+...f

(y) =

O+l

+...+n

(n:l)n

(“:‘).

And the general case is equivalent to Chapter 2’s rule

kn =

(n+l)m+’

integers

m,n

3 0,

Obk<n

m+l

’

if we divide both sides of this formula by m!. In fact, the addition formula

(5.8) tells us that

A((:))

(z’)-(iii)

(my’

if we replace

and k respectively by x + 1 and m. Hence the methods of

Chapter 2 give us the handy indefinite summation formula

L(z)”

(m;,)+”

162 BINOMIAL COEFFICIENTS

Binomial coefficients get their name from the binomial theorem, which

deals with powers of the binomial expression x + y. Let’s look at the smallest

cases of this theorem:

(x+y)O =

lxOyO

(x+y)' =

Ix'yO

lxc'y'

(x+y)Z =

lxZy0-t2x'y'

+lxOy2

(X+y)3 =

lx3yO

fSx2y' +3x'y2+1xOy3

(x+Y)~ =

1x4yo

+4x3y' +6x2y2 +4x'y3 +1x"y4.

It’s not hard to see why these coefficients are the same as the numbers in

Pascal’s triangle: When we expand the product

tX+t)n =

ix+Y)(x+Y)...b+d,

every term is itself the product of n factors, each either an x or

The number

of such terms with k factors of x and n

k factors of y is the coefficient

of xkyndk after we combine like terms. And this is exactly the number of

ways to choose k of the n binomials from which an x will be contributed; that

is, it’s (E).

Some textbooks leave the quantity

O”

undefined, because the functions

x0 and

0”

have different limiting values when x decreases to 0. But this is a

mistake. We must define

x0 = 1, for all x,

if the binomial theorem is to be valid when x = 0, y = 0, and/or x =

-y.

The theorem is too important to be arbitrarily restricted! By contrast, the

function

is quite unimportant.

But what exactly is the binomial theorem? In its full glory it is the

following identity:

(x + y)’ =

;

xky’--k,

integer

3 0

lx/y1

(5.12)

“At the age

of twenty-one

he [Moriarty] wrote

a treatise upon the

Binomial Theorem,

which has had a Eu-

ropean vogue. On

the strength of it,

he won the Math-

ematical Chair at

one of our smaller

Universities.”

-5’.

Holmes

1711

The sum is over all integers k; but it is really a finite sum when

is a nonneg-

ative integer, because all terms are zero except those with 0 6 k 6

On the

other hand, the theorem is also valid when

is negative, or even when

an arbitrary real or complex number. In such cases the sum really is infinite,

and we must have

ix/y1

< 1 to guarantee the sum’s absolute convergence.

5.1 BASIC IDENTITIES 163

Two special cases of the binomial theorem are worth special attention,

even though they are extremely simple. If x = y = 1 and

= n is nonnegative,

we get

2n =

(J+(y)+.-+(;),

integer n 3 0.

This equation tells us that row n of Pascal’s triangle sums to 2”. And when

x is -1 instead of

fl,

we get

0" =

(I)-(Y)+...+(-l)Q

integer n 3 0.

For example, 1

4 + 6

4 + 1 = 0; the elements of row n sum to zero if we

give them alternating signs, except in the top row (when n = 0 and

O”

= 1).

When

is not a nonnegative integer, we most often use the binomial

theorem in the special case y = 1. Let’s state this special case explicitly,

writing z instead of x to emphasize the fact that an arbitrary complex number

can be involved here:

+z)'

(;)z*,

IZI

< 1.

(5.13)

The general formula in (5.12) follows from this one if we set z = x/y and

multiply both sides by y’.

We have proved the binomial theorem only when

is a nonnegative in-

teger, by using a combinatorial interpretation. We can’t deduce the general

case from the nonnegative-integer case by using the polynomial argument,

because the sum is infinite in the general case. But when

is arbitrary, we

can use Taylor series and the theory of complex variables:

f"(0)

FZ2

+...

The derivatives of the function f(z) = (1 + z)’ are easily evaluated; in fact,

fckl(z)

(1 +

z)~~~.

Setting

= 0 gives (5.13).

(Chapter 9 tells the We also need to prove that the infinite sum converges, when

IzI

meaning of 0 .)

does, because (I) =

O(k-‘-‘)

by equation (5.83) below.

Now let’s look more closely at the values of (L) when n is a negative

integer. One way to approach these values is to use the addition law (5.8) to

fill in the entries that lie above the numbers in Table 155, thereby obtaining

Table 164. For example, we must have

(i’)

since (t) =

(i’)

+ (11) and

(1:) = 0; then we must have

(;‘)

= -1, since

(‘$

= (y’) +

(i’);

and so on.

164 BINOMIAL COEFFICIENTS

Table 164 Pascal’s triangle, extended upward.

n (a)

(7)

(I)

(t)

(a)

(:)

(i)

(‘d)

(;o)

-4 1 -4

-20

-56

-120

165

-220

286

-3 1 -3

-10

-21 28

-36

-55

-2 1 -2

3 -4. 5

-6

-8

-10

-1

1 -1

-1

0 0 0 0 0 0 0 0 0

All these numbers are familiar. Indeed, the rows and columns of Ta-

ble 164 appear as columns in Table 155 (but minus the minus signs). So

there must be a connection between the values of

(L)

for negative n and the

values for positive n. The general rule is

(-l)k(kp;-

‘)

, integer k;

it is easily proved, since

rk =

r(r-l)...(r-kkl)

(-l)k(-r)(l

-r)...(k-1

-r) = (-l)k(k-r-l)k

(5.14)

when k 3 0, and both sides are zero when k < 0.

Identity (5.14) is particularly valuable because it holds without any re-

striction. (Of course, the lower index must be an integer so that the binomial

coefficients are defined.) The transformation in (5.14) is called negating the

upper index, or “upper negation!’

But how can we remember this important formula? The other identities

we’ve seen-symmetry, absorption, addition, etc. -are pretty simple, but

this one looks rather messy. Still, there’s a mnemonic that’s not too bad: To

negate the upper index, we begin by writing down

(-l)k,

where k is the lower

index. (The lower index doesn’t change.) Then we immediately write k again,

twice, in both lower and upper index positions. Then we negate the original

upper index by subtracting it from the new upper index. And we complete

the job by subtracting 1 more (always subtracting, not adding, because this

is a negation process).

Let’s negate the upper index twice in succession, for practice. We get

(;) =

(-v(k-;-1)

You call this a

mnemonic? I’d call

pneumatic-

full

of air.

It does help me

remember, though.

(Now is a good

time to do warmup

exercise 4.)

(-1)2k

k-(k-r-l)-1

5.1 BASIC IDENTITIES 165

so we’re right back where we started. This is probably not what the framers of

R’s

also frustrating, the identity intended; but it’s reassuring to know that we haven’t gone astray.

if we’re

trying to

get somewhere else.

Some applications of (5.14) are, of course, more useful than this. We can

use upper negation, for example, to move quantities between upper and lower

index positions. The identity has a symmetric formulation,

(-I)-(-:

‘)

(-l)n(-mG

‘)

integers m,n 3 0,

(5.15)

which holds because both sides are equal to (“,‘“) .

Upper negation can also be used to derive the following interesting sum:

(5.16)

The idea is to negate the upper index, then apply (5.g), and negate again:

(Here double

nega-

tion helps, because

we’ve sandwiched

another operation in

between.)

(;)(-uk

(“-L-l)

kcm k$m

-r+m

(

(-l)m

+ml

(

This formula gives us a partial sum of the rth row of Pascal’s triangle, provided

that the entries of the row have been given alternating signs. For instance, if

r=5andm=2theformulagives1-5+10=6=(-1)2(~).

Notice that if m 3 r, (5.16) gives the alternating sum of the entire row,

and this sum is zero when r is a positive integer. We proved this before, when

we expanded (1

1)’ by the binomial theorem; it’s interesting to know that

the partial sums of this expression can also be evaluated in closed form.

How about the simpler partial sum,

L(L)

(I)

+..*+

(ii);

(5.17)

surely if we can evaluate the corresponding sum with alternating signs, we

ought to be able to do this one? But no; there is no closed form for the partial

sum of a row of Pascal’s triangle. We can do columns-that’s

(5.1o)-but

166 BINOMIAL COEFFICIENTS

not rows. Curiously, however, there is a way to partially sum the row elements

if they have been

multiplied1

by their distance from the center:

(I)

-k) =

Eq(m:

,),

integer m.

(5.18)

(This formula is easily verified by induction on m.) The relation between

these partial sums with and without the factor of (r/2

k) in the summand

is analogous to the relation between the integrals

xe+

+“.2

and

-m

c-i

-XLdx.

--oo

The apparently more compl.icated integral on the left, with the factor of x,

has a closed form, while the isimpler-looking integral on the right, without the

factor, has none. Appearances can be deceiving.

(Well,

it actually

At the end of this chapter, we’ll study a method by which it’s possible

equals

ifierf

to determine whether or not there is a closed form for the partial sums of a

a multiple of the

L‘err0r

f,,nction,,

given series involving binomial coefficients, in a fairly general setting. This

ifwe’re

will-

method is capable of discovering identities (5.16) and (5.18), and it also will

ing to accept that

tell us that (5.17) is a dead end.

as a closed form.)

Partial sums of the binomial series lead to a curious relationship of an-

other kind:

(mk+l)xkym-k

(J(-~)~(x+y)~~*,

integer m. (5.19)

k<m k<m

This identity isn’t hard to prove by induction: Both sides are zero when

m < 0 and 1 when m = 0. If we let

stand for the sum on the left, we can

apply the addition formula (5.8) and show easily that

‘m

&(m~~+r)Xkym~k+&(m~~~r)x~ym-k;

and

m-l

XkY

m-k

k<m

YSm-I

(m-i+r)Xm,

xsm-,

)

when m > 0. Hence

+y)SmpI

-z

(-X)”

(

5.1 BASIC IDENTITIES 167

and this recurrence is satisfied also by the right-hand side of (5.19). By

induction, both sides must be equal; QED.

But there’s a neater proof. When r is an integer in the range 0 3 r 3

-m,

the binomial theorem tells us that both sides of (5.19) are (x+y)“‘+‘y~‘. And

since both sides are polynomials in r of degree m or less, agreement at m + 1

different values is enough (but just barely!) to prove equality in general.

It may seem foolish to have an identity where one sum equals another.

Neither side is in closed form. But sometimes one side turns out to be easier

to evaluate than the other.

For example, if we set x = -1 and y =

we get

y(y)(-l,x

k<m

integer m 3 0,

an alternative form of identity (5.16). And if we set x = y = 1 and r = m +

we get

(‘“,’

‘)

(“:

“pk.

. .

The left-hand side sums just half of the binomial coefficients with upper index

2m +

and these are equal to their counterparts in the other half because

Pascal’s triangle has left-right symmetry.

Hence the left-hand side is just

1pm+1

22”

. This yields a formula that is quite unexpected,

(5.20)

Let’s check it when m = 2:

(‘,)

i(f)

i(i)

= 1 +

= 4. Astounding.

So far we’ve been looking either at binomial coefficients by themselves or

at sums of terms in which there’s only one binomial coefficient per term. But

many of the challenging problems we face involve products of two or more

binomial coefficients, so we’ll spend the rest of this section considering how

to deal with such cases.

Here’s a handy rule that often helps to simplify the product of two bino-

mial coefficients:

(L)(F)

(I)(z$

integers m, k.

(5.21)

We’ve already seen the special case k = 1; it’s the absorption identity (5.6).

Although both sides of (5.21) are products of binomial coefficients, one side

often is easier to sum because of interactions with the rest of a formula. For

example, the left side uses m twice, the right side uses it only once. Therefore

we usually want to replace

(i)

(r) by

(I;)

(A<“,)

when summing on

168 BINOMIAL COEFFICIENTS

Equation (5.21) holds primarily because of cancellation between m!‘s in

the factorial representations of (A) and (T) . If all variables are integers and

r 3 m 3 k 3 0, we have

r m

=--

m k m!(r-m)! k!(m-k)!

k! (m- k)! (r-m)!

(?.--I!

(;)(;;“k>.

k!(r-k)! (m-k)!(r-m)!

That was easy. Furthermore, if m < k or k < 0, both sides of (5.21) are

Yeah, right.

zero; so the identity holds for all integers m and k. Finally, the polynomial

argument extends its validity to all real r.

A binomial coefficient

1:;)

r!/(r

k)! k! can be written in the form

(a + b)!/a! b! after a suitab1.e renaming of variables. Similarly, the quantity

in the middle of the derivation above, r!/k! (m

k)! (r

m)!, can be written

in the form (a + b + ~)!/a! b! c!. This is a “trinomial coefficient

:’

which arises

in the “trinomial theorem” :

(x+y+z)n

(a+b+c)!

a! b! c!

xay

bZC

O$a,b,c<n

a+b+c=n

a+b+c

b+c

)(

xaybzc .

a+b+c=n

“Excogitavi autem

olim mirabilem

regulam pro

nu-

meris

coefficientibus

potestatum, non

So (A) (T) is really a trinomial coefficient in disguise. Trinomial coefficients

tanturn

bhomio

pop up occasionally in applications, and we can conveniently write them as

x + y , sed et a

trinomio

x + y +

imo a polynomio

quocunque,

data

potentia gradus

cujuscunque v.

(a + b + c)!

(aaTE,Tc)

a!b!

in order to emphasize the symmetry present.

gr.

decimi,

Binomial and trinomial coefficients generalize to multinomial coefi-

potentia in ejus

valore

comprehensa,

bents,

which are always expressible as products of binomial coefficients:

x5y3z2, possim

statim assignare

al + a2 + . . . + a,

(al

+az+...+a,)!

numerum coef-

ficientem, quem

al,a2,...,a,

al ! ar! . . . a,!

habere debet, sine

+ a2

. . . + a,

a2 + . . . + a,

“’

(““h,‘am)

ulla

Tabula

jam

calculata

--G.,V~~ibni~[~()fJ/

Therefore, when we run across such a beastie, our standard techniques apply.

5.1 BASIC IDENTITIES 169

Table 169 Sums of oroducts of binomial coefficients.

;

(m:3(*:k)

(SJ)

integers m, n. (5.22)

(,:,)

(n;k)

(,‘;;,)

1 integer

“”

(5.23)

integers m, n.

;

(m;k)

(“zk)(-lik

(-,)l+f;-;)

integer

13”

(5.24)

integers m, n.

(‘m”)

(k”n)(-l)k

(-l)L+m(;I;I;)

l,zy;o.

(5.25)

Now we come to Table 169, which lists identities that are among the most

important of our standard techniques. These are the ones we rely on when

struggling with a sum involving a product of two binomial coefficients. Each

of these identities is a sum over k, with one appearance of k in each binomial

coefficient; there also are four nearly independent parameters, called m, n,

etc., one in each index position. Different cases arise depending on whether k

appears in the upper or lower index, and on whether it appears with a plus or

minus sign. Sometimes there’s an additional factor of (-1

)k,

which is needed

to make the terms summable in closed form.

Fold

down the Table 169 is far too complicated to memorize in full; it is intended only

corner on this page,

so you can find the

for reference. But the first identity in this table is by far the most memorable,

table quickly later.

and it should be remembered. It states that the sum (over all integers k) of the

You’ll

need it!

product of two binomial coefficients, in which the upper indices are constant

and the lower indices have a constant sum for all k, is the binomial coefficient

obtained by summing both lower and upper indices. This identity is known

as Vandermonde’s convolution, because Alexandre Vandermonde wrote a

significant paper about it in the late 1700s

[293];

it was, however, known

to Chu Shih-Chieh in China as early as 1303. All of the other identities in

Table 169 can be obtained from Vandermonde’s convolution by doing things

like negating upper indices or applying the symmetry law, etc., with care;

therefore Vandermonde’s convolution is the most basic of all.

We can prove Vandermonde’s convolution by giving it a nice combinato-

rial interpretation. If we replace k by k

m and n by n

m, we can assume

170 BINOMIAL COEFFICIENTS

that m = 0; hence the identity to be proved is

(L)(nik)

(r:s)~

integer n.

(5.27)

Let

and s be nonnegative integers; the general case then follows by the

polynomial argument.

On the right side,

(‘L”)

is the number of ways to

choose n people from among r men and s women. On the left, each term

Sexist! You

men-

of the sum is the number of ways to choose k of the men and n

k of the

Coned men first.

women. Summing over all k. counts each possibility exactly once.

Much more often than n.ot we use these identities left to right, since that’s

the direction of simplification. But every once in a while it pays to go the

other direction, temporarily making an expression more complicated. When

this works, we’ve usually created a double sum for which we can interchange

the order of summation and then simplify.

Before moving on let’s look at proofs for two more of the identities in

Table 169. It’s easy to prove (5.23); all we need to do is replace the first

binomial coefficient by

(,-k-,),

then Vandermonde’s (5.22) applies.

The next one,

(5.24),

is a bit more difficult. We can reduce it to

Van-

dermonde’s convolution by a sequence of transformations, but we can just

as easily prove it by resorting to the old reliable technique of mathematical

induction. Induction is often the first thing to try when nothing else obvious

jumps out at us, and induction on

works just fine here.

For the basis

= 0, all terms are zero except when k =

-m;

so both sides

of the equation are (-l)m(s;m).

ow suppose that the identity holds for all

values less than some fixed

where

> 0. We can use the addition formula

replace

(,\,)

(,,!,yk)

i- (,i-,‘_,)

;

original sum now breaks into two

sums, each of which can be evaluated by the induction hypothesis:

(A,;)

(“‘I”)‘--‘)“+&

(m;;‘l)

(s;k)(-l)*

And this simplifies to the right-hand side of

(5.24),

if we apply the addition

formula once again.

Two things about this derivation are worthy of note. First, we see again

the great convenience of summing over all integers k, not just over a certain

range, because there’s no need to fuss over boundary conditions. Second,

the addition formula works nicely with mathematical induction, because it’s

a recurrence for binomial coefficients. A binomial coefficient whose upper

index is

is expressed in terms of two whose upper indices are

and

that’s exactly what we need to apply the induction hypothesis.

5.1 BASIC IDENTITIES 171

So much for Table 169. What about sums with three or more binomial

coefficients? If the index of summation is spread over all the coefficients, our

chances of finding a closed form aren’t great: Only a few closed forms are

known for sums of this kind, hence the sum we need might not match the

given specs. One of these rarities, proved in exercise 43, is

m n’

integers

m,n

3 0.

Here’s another, more symmetric example:

(a+b+c)!

a’b’c’

’

integers a, b, c 3 0.

. . .

(5.28)

(5.29)

This one has a two-coefficient counterpart,

~(~~~)(~:~)(-l)k

integersa,b>O,

(5.30)

which incidentally doesn’t appear in Table 169. The analogous four-coefficient

sum doesn’t have a closed form, but a similar sum does:

= (a+b+c+d)! (a+b+c)! (a+b+d)! (a+c+d)! (b+c+d)!

(2a+2b+2c+2d)!

(a+c)! (b+d)! a! b! c! d!

integers a, b, c, d 3 0.

This was discovered by John Dougall

[69]

early in the twentieth century.

Is Dougall’s identity the hairiest sum of binomial coefficients known? No!

The champion so far is

+...+a,

al,az,...,a,

integers

al,

al,.

. . , a, > 0.

(5.31)

Here the sum is over

(“r’)

index variables kii for 1 <

< j < n. Equation

(5.29) is the special case n = 3; the case n = 4 can be written out as follows,

172 BINOMIAL COEFFICIENTS

ifweuse (a,b,c,d) for

(al,az,as,Q)

and (i,j,k) for

(k12,k13,k23):

(a+b+c+d)!

a!b!c!d!

-’

integers a, b, c, d

The left side of (5.31) is the coefficient of 2:~;. . .zt after the product of

n(n

1) fractions

has been fully expanded into positive and negative powers of the

2’s.

The

right side of (5.31) was conjectured by Freeman Dyson in 1962 and proved by

several people shortly thereafter. Exercise 86 gives a “simple” proof of (5.31).

Another noteworthy identity involving lots of binomial coefficients is

;~-l)~+k(j;k)(;)(;)(m+;~~-k)

("n">

(:I;)

)

integers m, n > 0.

(5.32)

This one, proved in exercise 83, even has a chance of arising in practical

applications. But we’re getting far afield from our theme of “basic identities,’

so we had better stop and take stock of what we’ve learned.

We’ve seen that binomial coefficients satisfy an almost bewildering va-

riety of identities. Some of these, fortunately, are easily remembered, and

we can use the memorable ones to derive most of the others in a few steps.

Table 174 collects ten of the most useful formulas, all in one place; these are

the best identities to know.

5.2

BASIC PRA.CTICE

In the previous section we derived a bunch of identities by manipu-

lating sums and plugging in other identities. It wasn’t too tough to find those

derivations- we knew what we were trying to prove, so we could formulate

a general plan and fill in the details without much trouble. Usually, however,

out in the real world, we’re not faced with an identity to prove; we’re faced

with a sum to simplify. An.d we don’t know what a simplified form might

look like (or even if one exists). By tackling many such sums in this section

and the next, we will hone

clur

binomial coefficient tools.

Algorithm

self-teach:

1 read problem

2 attempt solution

skim book solu-

tion

4 ifattempt failed

&Ol

else

Rot0

problem

Unfortunately

that

algorithm

can put you in an

infinite loop.

Suggested patches:

&cc0

3a set c

c + 1

3b ifc = N

go& your TA

-E.

Dijkstra

But this

sub-

chapter is called

BASIC practice.

5.2 BASIC PRACTICE 173

To start, let’s try our hand at a few sums involving a single binomial

coefficient.

Problem 1: A sum of ratios.

We’d like to have a closed form for

(3/G)

)

integers n 3 m 3 0.

At first glance this sum evokes panic, because we haven’t seen any identi-

ties that deal with a quotient of binomial coefficients. (Furthermore the sum

involves two binomial coefficients, which seems to contradict the sentence

preceding this problem.) However, just as we can use the factorial represen-

tations to reexpress a product of binomial coefficients as another product

that’s how we got identity

(5.21)--e

can do likewise with a quotient. In

fact we can avoid the grubby factorial representations by letting r = n and

dividing both sides of equation (5.21) by

(i)

(t);

this yields

(T)/(L)

(Z)/(E).

So we replace the quotient on the left, which appears in our sum, by the one

on the right; the sum becomes

We still have a quotient, but the binomial coefficient in the denominator

doesn’t involve the index of summation k, so we can remove it from the sum.

We’ll restore it later.

We can also simplify the boundary conditions by summing over all k 3 0;

the terms for k > m are zero. The sum that’s left isn’t so intimidating:

(2)

It’s similar to the one in identity (5.g), because the index k appears twice

with the same sign. But here it’s -k and in (5.9) it’s not. The next step

should therefore be obvious; there’s only one reasonable thing to do:

(2)

174 BINOMIAL COEFFICIENTS

Table 174 The ton ten binomial coefficient identities.

=--

integers

k k!(n--k)!

’

nak>O.

factorial expansion

(E)

(n.l.k)

’

integer n 3 0,

integer k.

symmetry

integer k # 0.

absorption/extraction

(;)

(Ii’)

(;I:),

integer k.

addition/induction

(;)

(-l)k(kVL-‘),

integer k.

upper negation

integers m, k.

trinomial revision

integer

3 0,

Ix/y1

< 1.

binomial theorem

integer n.

parallel summation

integers

m,n>O.

upper summation

integer n. Vandermonde convolution

And now we can apply the parallel summation identity, (5.9):

n-mfk

‘(n-m) +m+ 1

k m

) =

(n;‘).

Finally’ we reinstate the

(k)

in the denominator that we removed from

the sum earlier, and then apply (5.7) to get the desired closed form:

(“;‘)/(:)

$A&*

This derivation actually works for any real value of n, as long as no division

by zero occurs; that is, as long as n isn’t one of the integers 0, 1, . . . , m

5.2 BASIC PRACTICE 175

The more complicated the derivation, the more important it is to check

the answer. This one wasn’t too complicated but we’ll check anyway. In the

small case m = 2 and n = 4 we have

(g/(40)

(f)/(Y)

($yJ

+i+i

yes, this agrees perfectly with our closed form (4 +

1)/(4

+ 1

2).

Problem 2: From the literature of sorting.

Our next sum appeared way back in ancient times (the early 1970s)

before people were fluent with binomial coefficients. A paper that introduced

an improved merging technique

[165]

concludes with the following remarks:

“It can be shown that the expected number of saved transfers . . is given by

the expression

Here m and n are as defined above, and mCn is the symbol for the number

of combinations of m objects taken n at a time. . . . The author is grateful to

the referee for reducing a more complex equation for expected transfers saved

to the form given here.”

Please, don’t

re-

mind me of the

midterm.

We’ll see that this is definitely not a final answer to the author’s problem.

It’s not even a midterm answer.

First we should translate the sum into something we can work with; the

ghastly notation

m-rPICm-n-l

is enough to stop anybody, save the enthusi-

astic referee (please). In our language we’d write

gk(zI:I:>/(:))

integers m > n 3 0.

The binomial coefficient in the denominator doesn’t involve the index of sum-

mation, so we can remove it and work with the new sum

What next? The index of summation appears in the upper index of the

binomial coefficient but not in the lower index. So if the other k weren’t there,

we could massage the sum and apply summation on the upper index (5.10).

With the extra k, though, we can’t. If we could somehow absorb that k into

the binomial coefficient, using one of our absorption identities, we could then

176 BINOMIAL COEFFICIENTS

sum on the upper index. Unfortunately those identities don’t work here. But

if the k were instead m

k, we could use absorption identity (5.6):

i--k)(~I~~~)

(m-n)(mmlE).

So here’s the key: We’ll rewrite k as m

k) and split the sum S

into two sums:

m-k-l

m-n-l

)

f(m-(m-kl)(~~~~:>

k=O

m-k-l

m-n-l

)

-f(m-ki(~I~~~)

k=O

(,“I:::)

-f(m-nJ(;g

k=O

mA-

(m-n)B,

where

The sums A and B that remain are none other than our old friends in

which the upper index varies while the lower index stays fixed. Let’s do B

first, because it looks simpler. A little bit of massaging is enough to make the

summand match the left side of (5.10):

In the last step we’ve included the terms with 0 6 k < m

n in the sum;

they’re all zero, because the upper index is less than the lower. Now we sum

on the upper index, using

(5.10),

and get

Do old exams

ever die?

5.2 BASIC PRACTICE 177

The other sum A is the same, but with m replaced by m

1. Hence we

have a closed form for the given sum S, which can be further simplified:

mA-(m-n)B

m(mmn)

-(m-n)(mrnn:,)

(m-Y+,)

(mmn)’

And this gives us a closed form for the original sum:

m-n+1

m-n

(

m-n+1

’

Even the referee can’t simplify this.

Again we use a small case to check the answer. When m = 4 and n = 2,

we have

ow(;)

lW@

24/(4,)

which agrees with our formula 2/(4

2 + 1).

Problem 3:

From

an old exam.

Let’s do one more sum that involves a single binomial coefficient. This

one, unlike the last, originated in the halls of academia; it was a problem on

a take-home test. We want the value of

Q~~OOOOO,

when

Qn =

(‘“k

‘)(-l)‘,

integer n 3 0.

k<2”

This one’s harder than the others; we can’t apply any of the identities we’ve

seen so far. And we’re faced with a sum of

2’oooooo

terms, so we can’t just

add them up. The index of summation k appears in both indices, upper and

lower, but with opposite signs. Negating the upper index doesn’t help, either;

it removes the factor of (-1

)k,

but it introduces a 2k in the upper index.

When nothing obvious works, we know that it’s best to look at small

cases. If we can’t spot a pattern and prove it by induction, at least we’ll have

178 BINOMIAL COEFFICIENTS

some data for checking our results. Here are the nonzero terms and their sums

for the first four values of

rt.

Qll

=1 =1

’

=1-l

(i)

(;)

(i)

= 1

= -1

@-((:)+($)-(;;)+(;)=l-7+15-lO+l=

We’d better not try the next case, n = 4; the chances of making an arithmetic

error are too high. (Computing terms like

(‘4’)

and (‘:) by hand, let alone

combining them with the others, is worthwhile only if we’re desperate.)

So the pattern starts out 1, 0, -1, 0. Even if we knew the next term or

two, the closed form wouldn’t be obvious. But if we could find and prove a

recurrence for

Q,,

we’d probably be able to guess and prove its closed form.

To find a recurrence, we need to relate

Q,--1

(or to

Qsmaiier

vaiues);

but

to do this we need to relate a term like

(12:J13),

which arises when n = 7 and

k = 13, to terms like

(“,;“).

This doesn’t look promising; we don’t know

any neat relations between entries in Pascal’s triangle that are 64 rows apart.

The addition formula, our main tool for induction proofs, only relates entries

that are one row apart.

But this leads us to a key observation: There’s no need to deal with

entries that are 2”-’ rows apart. The variable n never appears by itself, it’s

always in the context

2”.

So the

is a red herring! If we replace 2” by m,

Oh, the sneakiness

all we need to do is find a closed form for the more general (but easier) sum

of the instructor

who set that exam.

integer m 3 0;

then we’ll also have a closed form for

Q,,

Rz~.

And there’s a good chance

that the addition formula will give us a recurrence for the sequence R,.

Values of R, for small m can be read from Table 155, if we alternately

add and subtract values that appear in a southwest-to-northeast diagonal.

The results are:

There seems to be a lot of cancellation going on.

Let’s look now at the formula for R, and see if it defines a recurrence.

Our strategy is to apply the addition formula (5.8) and to find sums that

5.2 BASIC PRACTICE 179

have the form

in the resulting expression, somewhat as we did in the

perturbation method of Chapter 2:

m-l-k

)(-l)k

(m-;-k)(-)k+’

R,p,

(-1)‘”

R,p2

(-l)2(mp’i

R,e,

Rmp2.

Anyway those of

us who’ve done

warmup exercise 4

know it.

(In the next-to-last step we’ve used the formula (-,‘) =

(-l)“,

which we know

is true when m 3 0.) This derivation is valid for m 3 2.

From this recurrence we can generate values of

quickly, and we soon

perceive that the sequence is periodic. Indeed,

-1

if m mod 6 =

-1

The proof by induction is by inspection. Or, if we must give a more academic

proof, we can unfold the recurrence one step to obtain

= (R,p2

Rmp3)

R,-2

-Rm-3

whenever m 3 3. Hence

Rmp6

whenever m 3 6.

Finally, since Q,, =

Rzn,

we can determine Q,, by determining 2” mod 6

and using the closed form for

R,.

When n = 0 we have

mod 6 = 1; after

that we keep multiplying by 2 (mod 6), so the pattern 2, 4 repeats. Thus

{

=l,

ifn=O;

Q,, = Rp =

= 0, if n is odd;

R4=-I,

ifn>Oiseven.

This closed form for Qn agrees with the first four values we calculated when

we started on the problem. We conclude that

Q,OOOO~~

= -1.

180 BINOMIAL COEFFICIENTS

Problem 4: A sum involving two binomial coefficients.

Our next task is to

find:

a closed form for

integers m > n 3 0.

Wait a minute. Where’s the second binomial coefficient promised in the title

of this problem? And why should we try to simplify a sum we’ve already

simplified? (This is the sum S from Problem 2.)

Well, this is a sum that’s easier to simplify if we view the summand

as a product of two binomial coefficients, and then use one of the general

identities found in Table 169. The second binomial coefficient materializes

when we rewrite k as

(y):

And identity (5.26) is the one to apply, since its index of summation appears

in both upper indices and with opposite signs.

But our sum isn’t quite in the correct form yet. The upper limit of

summation should be m

1:)

if we’re to have a perfect match with (5.26). No

problem; the terms for n

k 6 m

1 are zero. So we can plug in, with

(I,

m,n, q)

m-n.

1,O);

the answer is

This is cleaner than the formula we got before. We can convert it to the

previous formula by using (5.7):

(m<+l)

= n ( m

)’m-n+1

m-n

Similarly, we can get interesting results by plugging special values into

the other general identities we’ve seen. Suppose, for example, that we set

m = n = 1 and q = 0 in (5.26). Then the identity reads

(l-k)k

(‘:‘).

O<k$l

Theleftsideis1((1+1)1/2)-(12+2’+..

L2),

so this gives us a brand new

way to solve the sum-of-squares problem that we beat to death in Chapter 2.

The moral of this story is: Special cases of very general sums are some-

times best handled in the general form. When learning general forms, it’s

wise to learn their simple specializations.

5.2 BASIC PRACTICE 181

Problem 5: A sum with three factors.

Here’s another sum that isn’t too bad. We wish to simplify

(ls)k,

integer n 3 0.

The index of summation k appears in both lower indices and with the same

sign; therefore identity (5.23) in Table 169 looks close to what we need. With

a bit of manipulation, we should be able to use it.

The biggest difference between (5.23) and what we have is the extra k in

our sum. But we can absorb k into one of the binomial coefficients by using

one of the absorption identities:

;

(;)

(2)s

(;)(;I:)

We don’t care that the s appears when the k disappears, because it’s constant.

And now we’re ready to apply the identity and get the closed form,

If we had chosen in the first step to absorb k into (L), not

(i),

we wouldn’t

have been allowed to apply (5.23) directly, because n

1 might be negative;

the identity requires a nonnegative value in at least one of the upper indices.

Problem 6: A sum with menacing characteristics.

The next sum is more challenging. We seek a closed form for

&(n:k’)rp)g,

integern30.

So we should

deep six this sum,

right?

One useful measure of a sum’s difficulty is the number of times the index of

summation appears. By this measure we’re in deep trouble-k appears six

times. Furthermore, the key step that worked in the previous problem-to

absorb something outside the binomial coefficients into one of them-won’t

work here. If we absorb the k + 1 we just get another occurrence of k in its

place. And not only that: Our index k is twice shackled with the coefficient 2

inside a binomial coefficient. Multiplicative constants are usually harder to

remove than additive constants.

182 BINOMIAL COEFFICIENTS

We’re lucky this time, though. The 2k’s are right where we need them

for identity (5.21) to apply, so we get

(“kk)

(T)k$

(TIk)

($3

The two 2’s disappear, and so does one occurrence of k. So that’s one down

and five to go.

The k+ 1 in the denominator is the most troublesome characteristic left,

and now we can absorb it into

(i)

using identity (5.6):

(Recall that n 3 0.) Two down, four to go.

To eliminate another k we have two promising options. We could use

symmetry on

(“lk);

or we could negate the upper index n + k, thereby elim-

inating that k as well as the factor

(-l)k.

Let’s explore both possibilities,

starting with the symmetry option:

(“:“)(;;:)(-‘Jk

(“n’“)(;++:)(-‘)*

Third down, three to go, and we’re in position to make a big gain by plugging

For a minute

into (5.24): Replacing (1, m, n, s) by (n + 1 ,

n, n), we get

thought we’d

have to punt.

Zero, eh? After all that work? Let’s check it when n = 2:

(‘,)

(i)

(i) (f)

(j)(i)+

= 1

= 0.

It checks.

Just for the heck of it, let’s explore our other option, negating the upper

index of

(“lk):

Now (5.23) applies, with

(l,m,n,s)

(n +

l,l,O,

-n

l),

and

hi;

(-nlF1)(z:)

s(t).

77~

binary search:

Replay the middle

formula first, to see

if the mistake was

early or late.

5.2 BASIC PRACTICE 183

Hey wait. This is zero when n > 0, but it’s 1 when n = 0. Our other

path to the solution told us that the sum was zero in all cases! What gives?

The sum actually does turn out to be 1 when n = 0, so the correct answer is

‘[n=O]‘. We must have made a mistake in the previous derivation.

Let’s do an instant replay on that derivation when n = 0, in order to see

where the discrepancy first arises. Ah yes; we fell into the old trap mentioned

earlier: We tried to apply symmetry when the upper index could be negative!

We were not justified in replacing (“lk) by (“zk) when k ranges over all

integers, because this converts zero into a nonzero value when k <

-n.

(Sorry

about that.)

The other factor in the sum, (L,‘:), turns out to be zero when k <

-n,

except when n = 0 and k = -1. Hence our error didn’t show up when we

checked the case n = 2. Exercise 6 explains what we should have done.

Problem 7: A new obstacle.

This one’s even tougher; we want a closed form for

integers

m,n

> 0.

If m were 0 we’d have the sum from the problem we just finished. But it’s

not, and we’re left with a real mess-nothing we used in Problem 6 works

here. (Especially not the crucial first step.)

However, if we could somehow get rid of the m, we could use the result

just derived. So our strategy is: Replace

(:Itk)

by a sum of terms like

(‘lt)

for some nonnegative integer

the summand will then look like the summand

in Problem 6, and we can interchange the order of summation.

What should we substitute for

(cztk)?

A painstaking examination of the

identities derived earlier in this chapter turns up only one suitable candidate,

namely equation (5.26) in Table 169. And one way to use it is to replace the

parameters

(L,

m, n,

k) by (n + k

1,2k, m

,O,

j), respectively:

(n+k2;l

-j)

(myl)

(2;)s

k>O O$j<n+k-1

&(mil)

,-z+,

(n+ki’-i)(T)%

‘k?O

In the last step we’ve changed the order of summation, manipulating the

conditions below the

1’s

according to the rules of Chapter 2.

184 BINOMIAL COEFFICIENTS

We can’t quite replace the inner sum using the result of Problem 6,

because it has the extra condition k > j

n + 1. But this extra condition

is superfluous unless j

n + 1 > 0; that is, unless j > n. And when j 3 n,

the first binomial coefficient of the inner sum is zero, because its upper index

is between 0 and k

1, thus strictly less than the lower index 2k. We may

therefore place the additional restriction j < n on the outer sum, without

affecting which nonzero terms are included. This makes the restriction k 3

n + 1 superfluous, and we can use the result of Problem 6. The double

sum now comes tumbling down:

I&)

~+k;l-i)~;)%

k>j-n+l

k>O

(,:,)In-1-j=O]

(:I:).

06j<n

The inner sums vanish except when j = n

so we get a simple closed form

as our answer.

Problem 8: A different obstacle.

Let’s branch out from Problem 6 in another way by considering the sum

&(n;k)(21;)k:;1:m’

integers

m,n

3 0.

Again, when m = 0 we have the sum we did before; but now the m occurs

in a different place. This problem is a bit harder yet than Problem 7, but

(fortunately) we’re getting better at finding solutions. We can begin as in

Problem 6,

Now (as in Problem 7) we try to expand the part that depends on m into

terms that we know how to deal with. When m was zero, we absorbed k + 1

into (z); if m > 0, we can do the same thing if we expand 1

/(k

+ 1 + m) into

absorbable terms. And our luck still holds: We proved a suitable identity

-1

r+l

integer m 3 0,

r+l-m’

7-g

{O,l,...,

m-l}.

(5.33)

5.2 BASIC PRACTICE 185

in Problem 1. Replacing

by -k

2 gives the desired expansion,

(“:“)

(1)&y&

(7)

(-k;2)~1.

Now the (k +

l)-’

can be absorbed into (z), as planned. In fact, it could

also be absorbed into (-kj- 2)p1. Double absorption suggests that even more

cancellation might be possible behind the scenes. Yes-expanding everything

in our new summand into factorials and going back to binomial coefficients

gives a formula that we can sum on k:

They expect

check this

on a

sheet of

sm =

(mE-t)!

j>.

~t-l)j(mn++;,+l)

(;;l++;;;)

(-n;

scratch paper.

m! n!

= (m+n+l)!

xc-

I.(

m+n+l

n+l+j n

’

j20

The sum over all integers j is zero, by (5.24). Hence

-S,

is the sum for j < 0.

To evaluate

-S,

for j < 0, let’s replace j by -k

1 and sum for k 3 0:

m! n!

= (m+n+l)!

k>O

~(-l)frn,+“k’l)

(-k;l)

. .

= (m+mnn+l)!

k<n

;lp,y-k(m+;+

‘>

(“n”-

‘>

m! n!

= (m+n+l)!

;:-,)*(m+;+l)

r;‘)

k<n

m! n!

(m+n+l)!

k<2n

,,,k(,,,+yy.

Finally (5.25) applies, and we have our answer:

sin

(-‘)n(my;;l)!

;

(-l)nm’l-mZ!d.,

Whew; we’d better check it. When n = 2 we find

s,=--

6 6

-+-

m(m- 1)

m+l

mS2

m+3 (m+l)(m+2)(m+3)

Our derivation requires m to be an integer, but the result holds for all real m,

because (m + 1

)n+'

is a polynomial in m of degree 6 n.

186 BINOMIAL COEFFICIENTS

5.3

TRICKS OF THE TRADE

Let’s look next at three techniques that significantly amplify the

methods we have already learned.

nick 1: Going halves.

This

should really

Many of our identities involve an arbitrary real number r. When

has

ca11ed

Trick

l/2

the special form “integer minus one half,” the binomial coefficient

can be

written as a quite different-looking product of binomial coefficients. This leads

to a new family of identities that can be manipulated with surprising ease.

One way to see how this works is to begin with the duplication formula

rk (r

5)” =

(2r)Zk/22k

)

integer k 3 0.

(5.34)

This identity is obvious if we expand the falling powers and interleave the

factors on the left side:

r(r--i)(r-l)(r-i)...(r-k+f)(r-k+i)

= (2r)(2r

1). . . (2r

2k+

2.2...:2

Now we can divide both sides by k!‘, and we get

(I;)

(y2)

(g/2”,

integer k.

(5.35)

If we set k =

= n, where n is an integer, this yields

integer n.

And negating the upper index gives yet another useful formula,

(-y2)

($)”

(:)

integer n.

For example, when n = 4 we have

(-l/2)(-3/2)(-5/2)(-7/2)

-1 2

)

1.2.3.4

1.3.5.7

-1

1.3.5.7.2.4.6.8

1.2.3.4.1.2.3.4

(;y(;).

(5.36)

(5.37)

we halve. .

Notice how we’ve changed a product of odd numbers into a factorial.

5.3 TRICKS OF THE TRADE 187

Identity

(5.35)

has

amusing

corollary.

Let

r =

in,

and take the

sum

over

all

integers

The

result is

(;k)

(2.32*

;

(y)

((y2)

n-1/2

(

17421

’

integer n 3

(5.33)

(5.23),

because

either

n/2

1)/2 is

Ln/2],

a nonnegative

integer!

can

also

use

Vandermonde’s

convolution

(5.27)

deduce

that

(-y’)

(R1/Zk) =

(:)

(-l)n,

integer n 3

Plugging in the values

from

(5.37)

gives

this is

what

sums

(-l)n.

Hence

have

remarkable property

the

“middle”

elements

Pascal’s

triangle:

&211)(2zIF)

= 4n,

integern>O.

(5.39)

For

example,

(z)

+($

(“,)+(“,)

(f)+($

(i) = 1.20+2.6+6.2+20.1 =

43.

These

illustrations

our

first

trick

indicate that

it’s

wise

try

changing

binomial

coefficients

the

form

(p) into binomial

coefficients

the

form

(nm;‘2),

where

n is

some

appropriate integer (usually

k);

the resulting

formula might

much

simpler.

Trick 2: High-order differences.

saw

earlier that

it’s

possible to evaluate partial

sums

the

series

(E)

(-1

)k,

but

not

the

series

(c).

turns

out

that

there

are

many important

applications

binomial

coefficients

with alternating

signs,

(t) (-1

)k.

One

the

reasons

for

this is that

such

coefficients

are

intimately

associated

with the

difference

operator

defined

Section

2.6.

The

difference

function

at the point

Af(x)

f(x

f(x)

;

188 BINOMIAL COEFFICIENTS

if we apply A again, we get the second difference

A2f(x) = Af(x + 1)

Af(x) = (f(x+Z)

f(x+l))

(f(x+l) -f(x))

= f(x+2)-2f(x+l)+f(x),

which is analogous to the second derivative. Similarly, we have

A3f(x) =

f(x+3)-3f(x+2)+3f(x+l)-f(x);

A4f(x)

f(x+4)-4f(x+3)+6f(x+2)-4f(x+l)+f(x);

and so on. Binomial coefficients enter these formulas with alternating signs.

In general, the nth difference is

A”f(x) =

(-l)"-kf(x+

k),

integer n 3

This formula is easily proved by induction, but there’s also a nice way to prove

it directly using the elementary theory of operators, Recall that Section 2.6

defines the shift operator E by the rule

Ef(x) = f(x+l);

hence the operator A is E

where 1 is the identity operator defined by the

rule

1 f(x)

f(x). By

the binomial theorem,

A” = (E-l)” =

(;)Ek(-l)"~k.

This is an equation whose elements are operators; it is equivalent to

(5.40)~

since Ek is the operator that takes

f(x)

into

f(x

k).

An interesting and important case arises when we consider negative

falling powers. Let f(x) = (x

1 )-’ = l/x. Then, by rule

(2.45),

we have

Af(x) = (-1)(x-

l)A,

A2f(x) = (-1)(-2)(x-

l)s,

and in general

A”((x-1)=1) =

(-1)%(x-l)*

= [-l)nx(X+l)n!.(x+n)

. .

Equation (5.40) now tells us that

x(x+l)...(x+n)

x+n

(

)

-1

’

x @{0,-l,...,

-n}.

(5.41)

5.3 TRICKS OF THE TRADE 189

For example,

1 4

---

x+1

f---

x+2

x+3

X x+4

x(x+1)(x+2)(x+3)(x+4)

l/x(xfi4).

The sum in (5.41) is the partial fraction expansion of n!/(x(x+l) . . . (x+n)).

Significant results can be obtained from positive falling powers too. If

f(x) is a polynomial of degree d, the difference Af(x) is a polynomial of degree

d-l

;

therefore

f(x) is a constant, and

f (x) = 0 if n > d. This extremely

important fact simplifies many formulas.

A closer look gives further information: Let

f(x)

adxd+ad~~xd-'+"'+a~x'+a~xo

be any polynomial of degree d. We will see in Chapter 6 that we can express

ordinary powers as sums of falling powers (for example, x2 =

+ xl); hence

there are coefficients

bd,

bdP1,

. . . ,

bl,

such that

f(X) =

bdX~+bd~,Xd-l+...+b,x~+box%

(It turns out that

and

ao,

but the intervening coefficients are

related in a more complicated way.) Let

= k!

for 0 6 k 6 d. Then

f(x)

= Cd(;)

+Cd-l(dy,)

+...+C,

(;>

.,(;)

;

thus, any polynomial can be represented as a sum of multiples of binomial

coefficients. Such an expansion is called the Newton series of f(x), because

Isaac Newton used it extensively.

We observed earlier in this chapter that the addition formula implies

‘((;))

(kr

Therefore, by induction, the nth difference of a Newton series is very simple:

A”f(X) =

(dxn)

‘cd&l(&~n)

““+‘l

(lTn)

+cO(Tn).

If we now set x = 0, all terms ck(kxn) on the right side are zero, except the

term with k-n = 0; hence

190 BINOMIAL COEFFICIENTS

The Newton series for f(x) is therefore

f(x)

= Adf(0)

;

+Ad-‘f(0)

+-.+.f,O,(;)

+f(O)(;)

For example, suppose f(x) = x3. It’s easy to calculate

f(0) = 0, f(1) = 1, f(2) = 8, f(3) = 27;

Af(0) = 1, Af(1) = 7,

Af(2) = 19;

A’f(0) = 6, A’f(1) = 12;

A3f(0) = 6.

So the Newton series is x3 =

6(:)

+6(l)

+ 1 (;) + O(i).

Our formula A” f(0) =

can also be stated in the following way, using

(5.40) with x = 0:

g;)(-uk(Co(~)+cl(;)+c2(~)+...)

= (-1)X,

integer n 3 0.

Here

(c~,cI,c~,...)

is an arbitrary sequence of coefficients; the infinite sum

co(~)+c,(:)+c2(:)+...

is actually finite for all k 3 0, so convergence is not

an issue. In particular, we can prove the important identity

(-l)k(ao+alk+...+a,kn)

(-l)%!a,,

integer n > 0,

(5.42)

because the polynomial

-t

k + . . . +

a,kn

can always be written as a

Newton series

CO(~)

(F)

-t . . . +

c,(E)

with

= n! a,.

Many sums that appear to be hopeless at first glance can actually be

summed almost trivially by using the idea of nth differences. For example,

let’s consider the identity

(‘n”“)

(-l)k

integer n > 0.

(5.43)

This looks very impressive, because it’s quite different from anything we’ve

seen so far. But it really is easy to understand, once we notice the telltale

factor

(c)(-l)k

in the summand, because the function

5.3 TRICKS OF THE TRADE 191

is a polynomial in k of degree n, with leading coefficient (-1 )“s”/n!. There-

fore (5.43) is nothing more than an application of (5.42).

We have discussed Newton series under the assumption that f(x) is a

polynomial. But we’ve also seen that infinite Newton series

f(x)

co(;)

+cl

(7)

+c2(;)

make sense too, because such sums are always finite when x is a nonnegative

integer. Our derivation of the formula A”f(0) =

c,,

works in the infinite case,

just as in the polynomial case; so we have the general identity

f(x)

f(O)(;)

+Af,O,(;)

.,f(O,(;)

+Ali(O,(;)

+...

integer x 3 0.

(5.44)

This formula is valid for any function f(x) that is defined for nonnegative

integers x. Moreover, if the right-hand side converges for other values of x,

it defines a function that “interpolates” f(x) in a natural way. (There are

infinitely many ways to interpolate function values, so we cannot assert that

(5.44) is true for all x that make the infinite series converge. For example,

if we let f(x) = sin(rrx), we have f(x) = 0 at all integer points, so the right-

hand side of (5.44) is identically zero; but the left-hand side is nonzero at all

noninteger x.)

A Newton series is finite calculus’s answer to infinite calculus’s Taylor

series. Just as a Taylor series can be written

9(a) s’(a) s”(a) 9”‘(a)

g(a+x) =

7X0

7X'

7x2+1x3

+...

(Since E = 1 + A, the Newton series for f(x) = g( a + x) can be written

E”

&(;)A”;

and

EXg(a)

s(a)

b(a)

A2 s(a)

g(a+x)

Tx”+Txl+T

x2 +

A3 s(a)

---x~+...

(5.45)

(This is the same as (5.44), because A”f(0) = A”g(a) for all n 3 0 when

f(x) = g( a + x).) Both the Taylor and Newton series are finite when g is a

polynomial, or when x = 0; in addition, the Newton series is finite when x is a

positive integer. Otherwise the sums may or may not converge for particular

values of x. If the Newton series converges when x is not a nonnegative integer,

it might actually converge to a value that’s different from g (a + x), because

the Newton series (5.45) depends only on the spaced-out function values g(a),

g(a +

l),

g(a + 2), . . . .

192 BINOMIAL COEFFICIENTS

One example of a convergent Newton series is provided by the binomial

theorem. Let g(x) = (1 + z)‘, where z is a fixed complex number such that

Iz/

< 1. Then Ag(x) = (1 + z) ‘+’

(1 + 2)’ =

~(1

+ z)‘, hence A”g(x) =

z”( 1 + 2)‘. In this case the infinite Newton series

g(a+X) =

tA”g(a)

+Z,“t

(;)zn

converges to the “correct” value (1 + z)“+‘, for all x.

James Stirling tried to use Newton series to generalize the factorial func-

tion to noninteger values. First he found coefficients S, such that

p(;)

so(;)

+s,(:>

+s2(;)

+...

(5.46)

is an identity for x = 0, x = 1, x = 2, etc. But he discovered that the resulting

“Forasmuch

series doesn’t converge except when x is a nonnegative integer. So he tried

these terms increase

again, this time writing

very fast, their

differences will

lnx! =

&h(z)

= SO(~)

+si(y)

+.2(i)

Now A(lnx!) = ln(x + l)!

lnx! = ln(x + l), hence

make a diverging

(5.47)

progression, which

hinders the

ordinate

of the parabola

from approaching to

the truth; therefore

in this and the like

An(ln41x=0

= A”-’ (ln(x + 1)) lxx0

cases,

interpolate

the logarithms of

the terms, whose

differences consti-

(-1 )n-‘Pk ln(k + 1)

tute a series swiftly

converging.

”

-J.

Stirling

12811

by (5.40). The coefficients are therefore

= 0;

= ln2;

= ln3

2 ln2 = In f;

ln4-3

ln3-t3 ln2 = In

$$;

etc. In this way Stirling obtained

(Proofs of

conver-

a series that does converge (although he didn’t prove it); in fact, his series

gence

were not

converges for all x > -1. He was thereby able to evaluate

satisfactorily.

invented until the

Exercise 88 tells the rest of the story.

nineteenth century.)

Trick 3: Inversion.

A special case of the rule (5.45) we’ve just derived for Newton’s series

can be rewritten in the following way:

d-4

(-llkfi’k)

f(n) =

(;)

(-l)kg(k). (5.48)

Znvert

this:

‘zmb

ppo’.

5.3 TRICKS OF THE TRADE 193

This dual relationship between f and g is called an inversion formula; it’s

rather like the Mobius inversion formulas (4.56) and (4.61) that we encoun-

tered in Chapter 4. Inversion formulas tell us how to solve “implicit recur-

rences,” where an unknown sequence is embedded in a sum.

For example, g(n) might be a known function, and f(n) might be

un-

known;andwemighthavefoundawaytoprovethatg(n)

=tk(t)(-l)kf(k).

Then (5.48) lets us express f(n) as a sum of known values.

We can prove (5.48) directly by using the basic methods at the beginning

of this chapter. If g(n) =

(T)(-l)kf(k)

for all n 3 0, then

(-1

)kg(k)

t-1

(r)

C-1

)‘f(i)

tfiii;

(11)1-ilk+‘(F)

xfij)&

G)(-llk+‘(~?)

~f(i,(~)

F(-l)*(nij)

[n-j=01

= f(n).

The proof in the other direction is, of course, the same, because the relation

between f and g is symmetric.

Let’s illustrate (5.48) by applying it to the “football victory problem”:

A group of n fans of the winning football team throw their hats high into the

air. The hats come back randomly, one hat to each of the n fans. How many

ways h(n, k) are there for exactly k fans to get their own hats back?

For example, if n = 4 and if the hats and fans are named A, B, C, D,

the

= 24 possible ways for hats to land generate the following numbers of

rightful owners:

ABCD 4 BACD 2 CABD

DABC

ABDC 2 BADC

CADB

DACB

ACBD 2 BCAD

CBAD 2 DBAC

ACDB

BCDA

CBDA

DBCA 2

ADBC

BDAC

CDAB

DCAB

ADCB 2 BDCA

CDBA

DCBA

Therefore h(4,4) = 1; h(4,3) = 0; h(4,2) = 6;

h(4,l)

= 8; h(4,O) = 9.

194 BINOMIAL COEFFICIENTS

We can determine h(n, k) by noticing that it is the number of ways to

choose k lucky hat owners, namely (L), times the number of ways to arrange

the remaining n-k hats so that none of them goes to the right owner, namely

h(n

k, 0). A permutation is called a derangement if it moves every item,

and the number of derangements of n objects is sometimes denoted by the

symbol ‘ni’, read “n subfactorial!’ Therefore h(n

k, 0) = (n

k)i, and we

have the general formula

h(n,k) =

(Subfactorial notation isn’t standard, and it’s not clearly a great idea; but

let’s try it awhile to see if we grow to like it. We can always resort to ‘D,’ or

something, if ‘ni’ doesn’t work out.)

Our problem would be solved if we had a closed form for ni, so let’s see

what we can find. There’s an easy way to get a recurrence, because the sum

of h(n, k) for all k is the total number of permutations of n hats:

n! =

xh(n,k)

($(n-k)i

k k

integer n 3 0.

(We’ve changed k to n

k and

(,“,)

(L)

in the last step.) With this

implicit recurrence we can compute all the h(n, k)‘s we like:

h(n,

h(n,2)

h(n,3)

h(n,4)

h(n,5)

h(n,

2 3 0

9 8 6

24645

10 0

135 40

15 0

For example, here’s how the row for n = 4 can be computed: The two right-

most entries are obvious-there’s just one way for all hats to land correctly,

and there’s no way for just three fans to get their own. (Whose hat would the

fourth fan get?) When k = 2 and k = 1, we can use our equation for h(n, k),

giving h(4,2) = ($h(2,0) = 6.1 = 6, and

h(4,l)

= (;)h(3,0) = 4.2 = 8. We

can’t use this equation for h(4,O); rather, we can, but it gives us h(4,O) =

(;)h(4,0),

h’

h . tw

is rue but useless. Taking another tack, we can use the

The art of math-

relation h(4,O) + 8 + 6 + 0 + 1 =

to deduce that h(4,O) = 9; this is the value

ematics, as of life,

is knowing which

of 4i. Similarly ni depends on the values of

for k < n.

truths are useless.

5.3 TRICKS OF THE TRADE 195

Baseball fans: .367

is also Ty

Cobb’s

lifetime batting

average, the a//-time

record.

Can this be

a coincidence?

(Hey wait, you’re

fudging.

Cobb

‘s

average was

4191/11429

.366699,

while

l/e

.367879.

But maybe if

Wade Boggs has

a few really good

seasons. . .

)

How can we solve a recurrence like (5.4g)? Easy; it has the form of (5.48),

with g(n) = n! and f(k) =

(-l)kki.

Hence its solution is

(-l)“t

Well, this isn’t really a solution; it’s a sum that should be put into closed form

if possible. But it’s better than a recurrence. The sum can be simplified, since

k! cancels with a hidden k! in

(i),

so let’s try that: We get

?li

n!il]“+k

(-‘lk

Oik<n

k)!

O<k<n k!

(5.50)

The remaining sum converges rapidly to the number

tkaO(-l

)k/k! =

e-l.

In fact, the terms that are excluded from the sum are

&!$?t(-,jk(;;n+:)i),

k20

(-l)n+’

, _ 1

=---

n+l

n+2

(n+2)l(n+3)

-“’

and the parenthesized quantity lies between 1 and 1

Therefore

the difference between ni and n!/e is roughly l/n in absolute value; more

precisely, it lies between 1

/(n

+ 1) and 1

/(n

+ 2). But

is an integer.

Therefore it must be what we get when we round n!/e to the nearest integer,

if n > 0. So we have the closed form we seek:

Tli

G+t

[n=O].

(5;51)

This is the number of ways that no fan gets the right hat back. When

n is large, it’s more meaningful to know the probability that this happens.

If we assume that each of the n! arrangements is equally likely- because the

hats were thrown extremely high- this probability is

n!/e + O(1)

;

.367..

So when n gets large the probability that all hats are misplaced is almost 37%.

Incidentally, recurrence (5.49) for subfactorials is exactly the same as

(5.46),

the firs recurrence considered by Stirling when he was trying to gen-t

eralize the factorial function. Hence

= ki. These coefficients are so large,

it’s no wonder the infinite series (5.46) diverges for noninteger x.

Before leaving this problem, let’s look briefly at two interesting patterns

that leap out at us in the table of small h(n, k). First, it seems that the num-

bers 1, 3, 6, 10, 15, . . . below the all-0 diagonal are the triangular numbers.

196 BINOMIAL COEFFICIENTS

This observation is easy to prove, since those table entries are the

h(n,n-2)‘s

and we have

h(n,n-2) =

(3,

It also seems that the numbers in the first two columns differ by fl. Is

this always true? Yes,

h(n,O)-h(n,l)

ni-n(n-l)i

n(n-l)!

O<k$n-1

n!(-‘)”

(-l)n

In other words, ni = n(n

l)l + (-1)“.

This is a much simpler recurrence

for the’ derangement numbers than we had before.

Now let’s invert something else. If we apply inversion to the formula

But inversion is the

source of smog.

that we derived in

(5.41),

we find

&(;):-li"(yp'.

x+n

This is interesting, but not really new. If we negate the upper index in (“lk),

we have merely discovered identity (5.33) again.

5.4 GENERATING FUNCTIONS

We come now to the most important idea in this whole book, the

notion of a generating function. An infinite sequence (Q,

al,

a~,

. . . ) that

we wish to deal with in some way can conveniently be represented as a power

series in an auxiliary variable

A(z) =

ac+a,z+a2z2+...

to@“.

k>O

(5.52)

It’s appropriate to use the letter z as the name of the auxiliary variable, be-

cause we’ll often be thinking of z as a complex number. The theory of complex

variables conventionally uses

‘z’

in its formulas; power series (a.k.a. analytic

functions or holomorphic functions) are central to that theory.

5.4 GENERATING FUNCTIONS 197

We will be seeing lots of generating functions in subsequent chapters.

Indeed, Chapter 7 is entirely devoted to them. Our present goal is simply to

introduce the basic concepts, and to demonstrate the relevance of generating

functions to the study of binomial coefficients.

A generating function is useful because it’s a single quantity that repre-

sents an entire infinite sequence. We can often solve problems by first setting

up one or more generating functions, then by fooling around with those func-

tions until we know a lot about them, and finally by looking again at the

coefficients. With a little bit of luck, we’ll know enough about the function

to understand what we need to know about its coefficients.

If A(z) is any power series

akzk,

we will find it convenient to write

[z”]A(z) = a,,;

(5.53)

in other words,

[z”]

A(z) denotes the coefficient of

Z”

in A(z).

Let A(z) be the generating function for

(00,

al,

az,. .

as in

(5.52),

and

let B(z) be the generating function for another sequence (bo, bl , bz , . . ,

Then

the product A(z) B (z) is the power series

(ao+alz+azz2+...)(bs+blz+b2z2+..~)

aobo

(aobl

+ albo)z +

(aobz

+ albl +

a2bo)z2

+ ...

;

the coefficient of

2”

in this product is

sob,, +

b,-1

. . .

anbO

$lkb,pl,.

k=O

Therefore if we wish to evaluate any sum that has the general form

akbn-k,

k=O

(5.54)

and if we know the generating functions A(z) and B(z) , we have

A(z)B(z)

The sequence (c,) defined by (5.54) is called the

conwo2ution

of the se-

quences (a,) and (b,); two sequences are “convolved” by forming the sums of

all products whose subscripts add up to a given amount. The gist of the previ-

ous paragraph is that convolution of sequences corresponds to multiplication

of their generating functions.

198 BINOMIAL COEFFICIENTS

Generating functions give us powerful ways to discover and/or prove

identities. For example, the binomial theorem tells us that (1 +

z)~

is the

generating function for the sequence

((i)

, (;) , (;) , . . ):

+z)'

(;)2

k30

Similarly,

+z)”

(;)zk.

k>O

If we multiply these

togethe:r,

we get another generating function:

+z)T(l

+z)S

= (1

+z)'+s.

And now comes the punch line: Equating coefficients of z” on both sides of

this equation gives us

g:)(A)

(T).

We’ve discovered Vandermonde’s convolution, (5.27)!

[5.27)!

That was nice and easy; let’s try another. This time we use (1

-z)~,

which

(5.27)[4.27)

is the generating function for the sequence

((-1

)"(G))

((h)

-(;),

(i)

, . . .

(3.27)[2.27)

Multiplying by (1 +

z)~

gives another generating function whose coefficients

(1.27)(0.27)!.

we know:

(1 --

z)'(l

+ z)' = (1

z2)'.

Equating coefficients of z” now gives the equation

~(~)(n~k)t-lik

(-1)n12(~,)Inevenl.

(5.55)

We should check this on a small case or two. When n = 3, for example,

the result is

(a)(;)-(F)(;)+(I)(T)-(;)(6)

Each positive term is cancelled by a corresponding negative term. And the

same thing happens whenever n is odd, in which case the sum isn’t very

5.4 GENERATING FUNCTIONS 199

interesting. But when n is even, say n = 2, we get a nontrivial sum that’s

different from Vandermonde’s convolution:

(ii)(;)-(;)(;)+(;)(;)

=2(i)-r’=

-?.

So (5.55) checks out fine when n = 2. It turns out that (5.30) is a special case

of our new identity (5.55).

Binomial coefficients also show up in some other generating functions,

most notably the following important identities in which the lower index

stays fixed and the upper index varies:

lfyou have a high-

lighter pen, these

(1 -Z)n+'

t(nn+k)zk,

integern30

k>O

two

equations

have

got to be marked.

integer n 3 0.

(5.56)

(5.57)

The second identity here is just the first one multiplied by

zn,

that is, “shifted

right” by n places. The first identity is just a special case of the binomial

theorem in slight disguise: If we expand (1

z)-~-’

by (5.13), the coefficient

of zk is

(-“,-‘)(-l)“,

which can be rewritten as (kl”) or (n:k) by negating

the upper index. These special cases are worth noting explicitly, because they

arise so frequently in applications.

When n = 0 we get a special case of a special case, the geometric series:

1-z

+z+z2

+z3

+ . . . =

X2".

k>O

This is the generating function for the sequence (1 , 1 ,

. . . ), and it is espe-

cially useful because the convolution of any other sequence with this one is

the sequence of sums: When

= 1 for all k, (5.54) reduces to

cn =

ak.

k=O

Therefore if A(z) is the generating function for the summands

(ao,

al , a2, . ),

then

A(z)/(l

-2)

is the generating function for the sums

(CO,CI

,cz,.

.).

The problem of derangements, which we solved by inversion in connection

with hats and football fans, can be resolved with generating functions in an

interesting way. The basic recurrence

n! =

(n-k)i

200 BINOMIAL COEFFICIENTS

can be put into the form of a convolution if we expand

(L)

in factorials and

divide both sides by n!:

(n-k)i

1=x-p.

k=O

k! (n-k)!

The generating function for the sequence (A,

. . . ) is

e’;

hence if we let

D(z) =

3zk,

k>O k!

the convolution/recurrence tells us that

= e’D(z).

1-z

Solving for D(z) gives

D(z) =

&eP

Equating coefficients of

2”

now tells us that

this is the formula we derived earlier by inversion.

So far our explorations with generating functions have given us slick

proofs of things that we already knew how to derive by more cumbersome

methods. But we haven’t used generating functions to obtain any new re-

sults, except for (5.55). Now we’re ready for something new and more sur-

prising. There are two families of power series that generate an especially rich

class of binomial coefficient identities: Let us define the generalized binomial

series

IBt

(z) and the generalized exponential series Et(z) as follows:

T&(z)

t(tk)*-‘;;

E,(z) =

t(tk+

l)k-’

(5.58)

k>O

It can be shown that these functions satisfy the identities

B,(z)‘-

-T&(z)-’

2;;

&t(z)-tln&t(z) = z.

(5.59)

In the special case t = 0, we have

730(z)

= 1

fz;

&O(Z)

= e’;

5.4 GENERATING FUNCTIONS 201

this explains why the series with parameter t are called “generalized” bino-

mials and exponentials.

The following pairs of identities are valid for all real r:

CBS,(z)’

(tk;

‘)

g-+zk;

k20

(5.60)

B,(zlr

1 -t+tcBt(z)

Et(z)’

-z&(z)

(tk+dkzk

(5.61)

k?O

’

(When tk + r = 0, we have to be a little careful about how the coefficient

is interpreted; each coefficient is a polynomial in r. For example, the

constant term of E,(z)~ is

r(0

r)-',

and this is equal to 1 even when r = 0.)

Since equations (5.60) and (5.61) hold for all r, we get very general iden-

tities when we multiply together the series that correspond to different powers

r and s. For example,

%(Zlr

%(zlS

1 -t+tBBt(z)

("l')

&,k

('j

s)zj

k20

gng

(‘“;r)-&)n;krs).

/ /

This power series must equal

IBt(Z)‘+S

-t+tt’B,(z)-’

n>O

tn+r+s n,

’

hence we can equate coefficients of zn and get the identity

(

t(:lkjiis)

tk&

(tn,.+s)

integer n,

valid for all real r, s, and t. When t = 0 this identity reduces to Vander-

monde’s convolution. (If by chance tk + r happens to equal zero in this

formula, the denominator factor tk + r should be considered to cancel with

the tk+r in the numerator of the binomial coefficient. Both sides of the iden-

tity are polynomials in r, s, and t.) Similar identities hold when we multiply

‘B,(z)’ by ‘B,(z)‘, etc.; Table 202 presents the results.

202 BINOMIAL COEFFICIENTS

Table 202 General convolution identities, valid for integer n 3 0.

= (tn+ r+s)ntnT++rS+S.

(5.65)

(5.62)

(5.63)

(5.64)

We have learned that it’s generally a good idea to look at special cases of

general results. What happens, for example, if we set t = l? The generalized

binomial

‘BI

(z) is very simple-it’s just

B,(z)

X2”

k>O

therefore

IB1

(z) doesn’t give us anything we didn’t already know from

Van-

dermonde’s convolution. But El (z) is an important function,

&(z)

x(k+,)k-l;

l+z+;~~+$r~+$~+...

(5.66)

k>O

that we haven’t seen before; it satisfies the basic identity Ah!

This is the

iterated power

&(z)

= ,=Q)

function

(5.67)

E(1n.z)

zLz’.

that I’ve often

This function, first studied by Eisenstein

[75],

arises in many applications.

wondered

about.

The special cases t = 2 and t = -1 of the generalized binomial are of

zztrzr,,

particular interest, because their coefficients occur again and again in prob-

lems that have a recursive structure. Therefore it’s useful to display these

5.4 GENERATING FUNCTIONS

2~1

series explicitly for future reference:

.qy)&

1-y.

(5.68)

(5%)

(5.70)

(5.71)

(5.72)

(5.73)

The coefficients

(y)

(z) are called the Catalan numbers C,, because

Eugene Catalan wrote an influential paper about them in the 1830s

[46].

The

sequence begins as follows:

n 0’2345

6 7 8 9

1 1 2 5 14 42

‘32 429 ‘430

4862 ‘6796

The coefficients of B-1 (z) are essentially the same, but there’s an extra 1 at the

beginning and the other numbers alternate in sign: (1,

-1,2,

-5,14,.

. .

Thus

BP1

(z) = 1 +

zBz(-z).

We also have

1 (z) = %2(-z)

‘.

Let’s ClOSe this section by deriving an important consequence of (5.72)

and (5.73), a relation that shows further connections between the functions

L!L, (z) and ‘Bz(-z):

B-1

(z)n+’

(-Z)n+‘B~(-Z)n+’

VTFG

(yk)z,

k<n

204 BINOMIAL COEFFICIENTS

This holds because the coefficient of

(-z)“+“B2(-~)“~‘/~~

(-,)n+l[Zk

n-11

(-1

)n+l(-,

)km

n 1

[Zkmnpl]

B2(Z)n+’

dixz

= (-1y

2(k-n-l)+n+l

k--n- 1

(-l)k

r;I;I-;)

(-,)k('"-;-')

n-k

)

,z”,

%-I

(Z)n+’

JiTz

when k > n. The terms nicely cancel each other out. We can now use (5.68)

and (5.69) to obtain the closed form

integer n > 0.

(5.74)

(The special case

= -1 came up in Problem 3 of Section 5.2. Since the

numbers

$(l

are sixth roots of unity, the sums

tks,,

(“ik)(-l)k

have the periodic behavior we observed in that problem.) Similarly we can

combine (5.70) with (5.71) to cancel the large coefficients and get

(l+yG)‘+(l-ywz)y

integer n > 0.

(5.75)

5.5

HYPERGEOMETRIC FUNCTIONS

The methods we’ve been applying to binomial coefficients are very

effective, when they work, but we must admit that they often appear to be

ad hoc-more like tricks than techniques. When we’re working on a problem,

we often have many directions to pursue, and we might find ourselves going

They’re even more

around in circles. Binomial coefficients are like chameleons, changing their

versatile than

appearance easily. Therefore it’s natural to ask if there isn’t some unifying

chameleons; we

can dissect them

principle that will systematically handle a great variety of binomial coefficient

and

put

them

summations all at once. Fortunately, the answer is yes. The unifying principle

back together in

is based on the theory of certain infinite sums called hypergeometric series.

different ways.

5.5 HYPERGEOMETRIC FUNCTIONS 205

The study of hypergeometric series was launched many years ago by Eu-

ler, Gauss, and Riemann; such series, in fact, are still the subject of consid-

erable research. But hypergeometrics have a somewhat formidable notation,

Anything that has

which takes a little time to get used to.

survived for cen-

turies with such

The general hypergeometric series is a power series in

with m + n

awesome

notation

parameters, and it is defined as follows in terms of rising factorial powers:

must be really

useful.

(

al,

..',

aIlI

i; k

a’

...am

bl,

.-.,bn

’

k>O

by.

. .

(5.76)

To avoid division by zero, none of the b’s may be zero or a negative integer.

Other than that, the a’s and b’s may be anything we like. The notation

‘F(al,.

. . ,a,,,;

bl,.

. ,

b,;

z)’

is also used as an alternative to the two-line form

(5.76),

since a one-line form sometimes works better typographically. The a’s

are said to be upper parameters; they occur in the numerator of the terms

of F. The b’s are lower parameters, and they occur in the denominator. The

final quantity

is called the argument.

Standard reference books often use

’

,,,F,’ instead of ‘F’ as the name of a

hypergeometric with m upper parameters and n lower parameters. But the

extra subscripts tend to clutter up the formulas and waste our time, if we’re

compelled to write them over and over. We can count how many parameters

there are, so we usually don’t need extra additional unnecessary redundancy.

Many important functions occur as special cases of the general hypergeo-

metric; indeed, that’s why hypergeometrics are so powerful. For example, the

simplest case occurs when m = n = 0: There are no parameters at all, and

we get the familiar series

1~)

= e’.

Actually the notation looks a bit unsettling when m or n is zero. We can add

an extra ‘1’ above and below in order to avoid this:

In general we don’t change the function if we cancel a parameter that occurs

in both numerator and denominator, or if we insert two identical parameters.

The next simplest case has m = 1, al = 1, and n = 0; we change the

parameterstom=2,

=al=l,

n=l,andbl

=l,sothatn>O.

This

series also turns out to be familiar, because

1’

= k!:

206 BINOMIAL COEFFICIENTS

It’s our old friend, the geometric series; F( a’, . . . , a,,,; b’ , . . . ,

b,;

z) is called

hypergeometric because it includes the geometric series F(

1,l;

1; z) as a very

special case.

The general case m = 1 and n = 0 is, in fact, easy to sum in closed form,

La';

~(a'~p')zk

-z)(l

’

(5.77)

k20

using (5.56). If we replace a by -a and

-2,

we get the binomial theorem,

F(-4

1-z)

= (l+z)"

A negative integer as upper parameter causes the infinite series to become

finite, since (-a)” = 0 whenever k > a 3 0 and a is an integer.

The general case m = 0, n = 1 is another famous series, but it’s not as

well known in the literature of discrete mathematics:

(5.78)

This function

I’,

’ is called a “modified Bessel function” of order b

1. The

special case b = 1 gives us F( ,‘, lz) =

10(2&),

which is the interesting series

k20

zk/k!‘.

The special case m = n = 1 is called a “confluent hypergeometric series”

and often denoted by the letter M:

ak zk

M(a,b,z)

k>O

(5.79)

This function, which has important applications to engineering, was intro-

duced by Ernst Kummer.

By now a few of us are wondering why we haven’t discussed convergence

of the infinite series (5.76). The answer is that we can ignore convergence if

we are using

simply as a formal symbol. It is not difficult to verify that

formal infinite sums of the form

tk3,,

(Xkzk

form a field, if the coefficients

ak lie in a field. We can add, subtract, multiply, divide, differentiate, and do

functional composition on such formal sums without worrying about conver-

gence; any identities we derive will still be formally true. For example, the

hypergeometric F( “i

,’

/z) = tkZO k!

doesn’t converge for any nonzero z;

yet we’ll see in Chapter 7 that we can still use it to solve problems. On the

other hand, whenever we replace z by a particular numerical value, we do

have to be sure that the infinite sum is well defined.

5.5 HYPERGEOMETRIC FUNCTIONS 207

“There must

many universities

to-day

where 95

per cent, if not

100 per cent, of the

functions studied by

physics, engineering,

and even mathe-

matics

students,

are covered by

this single symbol

F(a,b;c;x).”

W. W. Sawyer[257]

The next step up in complication is actually the most famous

hypergeo-

metric of all. In fact, it was the hypergeometric series until about 1870, when

everything was generalized to arbitrary m and n. This one has two upper

parameters and one lower parameter:

a,b

(

akbk

/=t---.

k>O

ci;k!

(5.80)

It is often called the Gaussian hypergeometric, because many of its subtle

properties were first proved by Gauss in his doctoral dissertation of 1812

[116],

although Euler

[95]

and Pfaff

12331

had already discovered some remarkable

things about it. One of its important special cases is

k! k!

(-z)~

.zt-----

k>O

(k+

l)!

z--+--T+“’

2 3

Notice that

ZC’

ln( 1 +z) is a hypergeometric function, but ln( 1 +z) itself cannot

be hypergeometric, since a hypergeometric series always has the value 1 when

So far hypergeometrics haven’t actually done anything for us except pro-

vide an excuse for name-dropping. But we’ve seen that several very different

functions can all be regarded as hypergeometric; this will be the main point of

interest in what follows. We’ll see that a large class of sums can be written as

hypergeometric series in a “canonical” way, hence we will have a good filing

system for facts about binomial coefficients.

What series are hypergeometric? It’s easy to answer this question if we

look at the ratio between consecutive terms:

The first term is

and the other terms have ratios given by

fk+l

k+l

. ..a.

b:...bf:

Zk+l

_____

al . ..a.

bki’

, . . .

bk,+‘(k+l)!

(k+al)...(k+a,)z

(k+bl)...(k+b,)(k+l)’

This is a rational function of k, that is, a quotient of polynomials in k. Any

rational function of k can be factored over the complex numbers and put

208 BINOMIAL COEFFICIENTS

into this form. The a’s are the negatives of the roots of the polynomial in

the numerator, and the b’s are the negatives of the roots of the polynomial

in the denominator. If the denominator doesn’t already contain the special

factor (k + 1

we can include (k + 1) in both numerator and denominator. A

constant factor remains, and we can call it

Therefore hypergeometric series

are precisely those series whose first term is 1 and whose term ratio tk+l/tk

is a rational function of k.

Suppose, for example, that we’re given an infinite series with term ratio

tk+

k2+7k+10

4k2

’

a rational function of k. The numerator polynomial splits nicely into two

factors, (k + 2) (k +

5),

and the denominator is 4(k + i/2) (k

i/2). Since the

denominator is missing the required factor

(kf

l), we write the term ratio as

tk+

(k+2)(k+5)(k+

1)(1/4)

(k+i/2)(k-i/2)(k+

’

and we can read off the results: The given series is

k>O

toF(i;,?;2/V4).

Thus, we have a general method for finding the hypergeometric represen-

tation of a given quantity S, when such a representation is possible: First we

write S as an infinite series whose first term is

nonzero.

We choose a notation

so that the series is

k20

with

# 0. Then we

Cahhte

tk+l/tk. If the (NOW is a good

term ratio is not a rational function of k, we’re out of luck. Otherwise we

time

warmuP

express it in the form (5.81); this gives parameters al, . . . , a,, br, . . . , b,,

exercise 11.)

and an argument z, such that S =

al,.

. . , a,,,; br , . . . ,

b,;

z).

Gauss’s hypergeometric series can be written in the recursively factored

form

a+2

b+2

--z(1

+...)

3 c-t2

if we wish to emphasize the importance of term ratios.

Let’s try now to reformulate the binomial coefficient identities derived

earlier in this chapter, expressing them as hypergeometrics. For example,

let’s figure out what the parallel summation law,

&(‘i”>

(r,,+‘),

integern,

5.5 HYPERGEOMETRIC FUNCTIONS 209

looks like in hypergeometric notation. We need to write the sum as an infinite

series that starts at k = 0, so we replace k by n

r+n-k

n-k

(r+n-k)!

k,O

k)!

k>O

This series is formally infinite but actually finite, because the (n

k)! in the

denominator will make

= 0 when k > n. (We’ll see later that l/x! is

defined for all x, and that l/x! = 0 when x is a negative integer. But for now,

let’s blithely disregard such technicalities until we gain more hypergeometric

experience.) The term ratio is

tk+l

(r+n-k-l)!r!(n-k)!

n-k

= r!(n-k-l)!(r+n-k)! =

r+n-k

(k+

l)(k-n)(l)

= (k-n-r)(k+ 1)

Furthermore

= (“,“). Hence the parallel summation law is equivalent to

the hypergeometric identity

("n")r(:l+il)

(r+,,').

Dividing through by (“,“) g’

Ives

a slightly simpler version,

(5.82)

Let’s do another one. The term ratio of identity

(5.16),

integer

(k-m)/(r-m+k+l)

=(k+l)(k-m)(l)/(k-m+r+l)(k+l),

after

we replace k by m

k; hence (5.16) gives a closed form for

This is essentially the same as the hypergeometric function on the left of

(5.82),

but with m in place of n and r + 1 in place of

-r.

Therefore identity

(5.16) could have been derived from (5.82), the hypergeometric version of

(5.9). (No wonder we found it easy to prove (5.16) by using (5.g).)

First derangements,

Before we go further, we should think about degenerate cases, because

now degenerates.

hypergeometrics are not defined when a lower parameter is zero or a negative

210 BINOMIAL COEFFICIENTS

integer. We usually apply the parallel summation identity when

and n are

positive integers; but then

-n--r

is a negative integer and the hypergeometric

(5.76) is undefined. How

th.en

can we consider (5.82) to be legitimate? The

answer is that we can take the limit of F(

Pr,{TFE

11)

We will look at such things more closely later in this chapter, but for now

let’s just be aware that some denominators can be dynamite. It is interesting,

however, that the very first sum we’ve tried to express hypergeometrically

has turned out to be degenerate.

Another possibly sore point in our derivation of (5.82) is that we ex-

panded (“‘,“i”) as (r + n

k)!/r!

k)!. This expansion fails when r is a

negative integer, because (--m)! has to be m if the law

O! =

O.(-l).(-2)...:(-m+l).(-m)!

is going to hold. Again, we need to approach integer results by considering a

limit of

-4 0.

But we defined the factorial representation

(L)

r!/k!

(r-k)! only when

is an integer! If we want to work effectively with hypergeometrics, we need

a factorial function that is defined for all complex numbers. Fortunately there

is such a function, and it can be defined in many ways. Here’s one of the most

useful definitions of z!, actually a definition of 1 /z! :

lim

+’

‘.

n-03

(

)

(5.83)

(See exercise 21. Euler

[81]

discovered this when he was 22 years old.) The

limit can be shown to exist for all complex z, and it is zero only when

is a

negative integer. Another significant definition is

z! =

t’e

dt , if

312

> -1.

This integral exists only when the real part of z exceeds -1, but we can use

the formula

z! = z(z-l)!

(5.85)

to extend (5.84) to all complex z (except negative integers). Still another

definition comes from Stirl:ing’s interpolation of lnz! in (5.47). All of these

approaches lead to the same generalized factorial function.

There’s a very similar function called the Gamma function, which re-

lates to ordinary factorials somewhat as rising powers relate to falling powers.

Standard reference books often use factorials and Gamma functions simulta-

neously, and it’s convenient to convert between them if necessary using the

(We proved the

identities originally

for integer

and

used the polynomial

argument to show

that they hold in

general. Now we’re

proving them first

for irrational

and using a limiting

argument to show

that they ho/d for

integers!)

5.5 HYPERGEOMETRIC FUNCTIONS 211

following formulas:

How do you write

to the

power,

when W is the

complex conjugate

I see, the lower

index

arrives at

its limit first.

That’s why

(;)

is zero when

a negative integer.

T(z+l) =

z!;

(5.86)

(-z)!

T(z)

-T-.

sin

712

(5.87)

We can use these generalized factorials to define generalized factorial

powers, when

and w are arbitrary complex numbers:

(z-w)!

’

ryz

r(z)

The only proviso is that we must use appropriate limiting values when these

formulas give

CXI/OO.

(The formulas never give O/O, because factorials and

Gamma-function values are never zero.) A binomial coefficient can be written

= lim lim

W L-+2 w-w w! (<

w)!

(5.90)

when

and w are any complex numbers whatever.

Armed with generalized factorial tools, we can return to our goal of re-

ducing the identities derived earlier to their hypergeometric essences. The

binomial theorem (5.13) turns out to be neither more nor less than (5.77),

as we might expect. So the next most interesting identity to try is Vander-

monde’s convolution (5.27):

$)(n”k)

(‘i”)~

integer n.

The kth term here is

= (r-k)!k! (s-n+k)!(n-k)!

’

and we are no longer too shy to use generalized factorials in these expres-

sions. Whenever

contains a factor like

(LX

+ k)!, with a plus sign before

the k, we get (o1+ k +

l)!/(a

+ k)! = k + a + 1 in the term ratio

tk+j/tk,

by (5.85); this contributes the parameter

‘a+

1’ to the corresponding hyper-

geometric-as an upper parameter if (

+ k)! was in the numerator of tk,

but as a lower parameter otherwise. Similarly, a factor like

(LX

k)! leads to

l)!/(a

k)! =

(-l)/(k

a); this contributes ‘-a’ to the opposite

set of parameters (reversing the roles of upper and lower), and negates the

hypergeometric argument. Factors like r!, which are independent of k, go

212 BINOMIAL COEFFICIENTS

into

but disappear from

t,he

term ratio. Using such tricks we can predict

without further calculation

t;hat

the term ratio of (5.27) is

tk+l

k-r k

-=-

k+l

k+s-n+l

times

(--1

)’ =

and Vandermonde’s convolution becomes

(5.91)

We can use this equation to determine F( a, b; c; z) in general, when z = 1 and

when b is a negative integer.

Let’s rewrite (5.91) in a form so that table lookup is easy when a new

sum needs to be evaluated. The result turns out to be

a,b

(

, _

T(c-a--b)T(c)

r(c

T(c

’

integer b 6 0

>Ra+!Xb.

(5.92)

Vandermonde’s convolution (5.27) covers only the case that one of the upper

parameters, say b, is a nonpositive integer; but Gauss proved that (5.92) is

A few weeks ago, we

valid also when a, b, c are complex numbers whose real parts satisfy

!Xc

were studying what

%a +

%b.

In other cases, the infinite series F(

“;”

j 1) doesn’t converge. When

b =

-n,

the identity can be written more conveniently with factorial powers

~~~r~~r~e~e

Now

we’re

studying

instead of Gamma functions:

stuff beyond his

Ph.D. thesis.

F(a’;ni,)

k&z

(;-;s,

integer n > 0.

(5.93)

this intimidating

or what?

It turns out that all five of the identities in Table 169 are special cases of

Vandermonde’s convolution; formula (5.93) covers them all, when proper at-

tention is paid to degenerate situations.

Notice that (5.82) is just the special case a = 1 of (5.93). Therefore we

don’t really need to remember (5.82); and we don’t really need the identity

(5.9) that led us to (5.82), even though Table 174 said that it was memo-

rable. A computer program for formula manipulation, faced with the prob-

lem of evaluating

xkGn

(

‘+kk),

could convert the sum to a hypergeometric and

plug into the general identity for Vandermonde’s convolution.

Problem 1 in Section 5.2 asked for the value of

This problem is a natural for hypergeometrics, and after a bit of practice any

hypergeometer can read off the parameters immediately as F(

-m;

-n;

1).

Hmmm; that problem was yet another special takeoff on Vandermonde!

5.5 HYPERGEOMETRIC FUNCTIONS 213

The sum in Problem 2 and Problem 4 likewise yields F( 2,1

n; 2

m; 1).

(We need to replace k by k + 1 first.) And the “menacing” sum in Problem 6

turns out to be just F(n +

-n;

2; 1). Is there nothing more to sum, besides

disguised versions of Vandermonde’s powerful convolution?

Well, yes, Problem 3 is a bit different. It deals with a special case of the

general sum

(“kk) zk considered in (5.74), and this leads to a closed-form

expression for

We also proved something new in (5.55), when we looked at the coeffi-

cients of (1

z)~(

1 +

z)~:

l-c-2n,

-2n

(

(2n)!

-1

(-l)n-

1 )!

n! (c+n-l)!’

integer n 3 0.

Kummer was a

summer.

This is called Kummer’s formula when it’s generalized to complex numbers:

(5.94)

The summer of ‘36.

(Ernst Kummer

[187]

proved this in 1836.)

It’s interesting to compare these two formulas. Replacing c by

l -2n-

we find that the results are consistent if and only if

(5.95)

when n is a positive integer. Suppose, for example, that n = 3; then we

should have

-6!/3!

= lim

X+

3x!/(2x)!.

We know that

(-3)!

and

(-6)!

are

both infinite; but we might choose to ignore that difficulty and to imagine

that

(-3)!

(-3)(-4)(-5)(-6)!,

so that the two occurrences of

(-6)!

will

cancel. Such temptations must, however, be resisted, because they lead to

the wrong answer! The limit of x!/(2x)! as x

-3 is not (-3) (-4) (-5) but

rather

-6!/3!

(-4)(-5)(-6),

according to (5.95).

The right way to evaluate the limit in (5.95) is to use equation (5.87),

which relates negative-argument factorials to positive-argument Gamma func-

tions. If we replace x by -n +

and let

0, two applications of (5.87)

give

(-n-e)!

F(n+e)

sin(2n + 2e)rt

(-2n

2e)! F(2n +

2e)

sin(n + e)rc

214 BINOMIAL COEFFICIENTS

Now sin( x + y ) = sin x cos y + cos x sin y

;

so this ratio of sines is

cos

2n7t

sin

cos n7t sin

c7r

(-qn(2

+ O(e)) ,

by the methods of Chapter 9. Therefore, by (5.86), we have

!‘_mo

(-2n

2e)!

(-n-4!

2(-l),r(2n)

,(-,),P-l)!

Vn)!

r(n)

(n-l)!

(-‘)

7’

as desired.

Let’s complete our survey by restating the other identities we’ve seen so

far in this chapter, clothing them in hypergeometric garb. The triple-binomial

sum in (5.29) can be written

--a-2n,

-b-211,

-2n ,

(2n)!

(a+b+2n-2)”

(-l)nn!-

ak’,‘i

’

integer n 3 0.

When this one is generalized to complex numbers, it is called Dixon’s for-

mula:

= (c/2)!

(c-a)*(c-b)*

1 fc-a, 1

fc-b

(c-a-b)*

’

b6)

fla+Rb

1 +Rc/2.

One of the most general formulas we’ve encountered is the triple-binomial

sum (5.28), which yields Saalschiitz’s identity:

a, b,

--n

(c-a)K(c-b)”

afb-c-n+1

c”(c-a-b)K

c)n (b

c)E

(-c)s(a+b-c)n’

integer n 3 0.

This formula gives the value at

= 1 of the general hypergeometric series

with three upper parameters and two lower parameters, provided that one

of the upper parameters is a nonpositive integer and that

+ 1. (If the sum of the lower parameters exceeds the sum of the

upper parameters by 2 instead of by

the formula of exercise 25 can be used

to express F(al , a2, as;

, b2; 1) in terms of two hypergeometrics that satisfy

Saalschiitz’s identity.)

Our hard-won identity in Problem 8 of Section 5.2 reduces to

---F

(

x+1,

n+l,

-n

1+x

x+2

(-‘)nX”X-n=l.

5.5 HYPERGEOMETRIC FUNCTIONS 215

(Historical note:

The great relevance

of hypergeometric

series to binomial

coefficient identities

was first pointed

out by George

Andrews in 1974

/9,

section

51.)

Sigh. This is just the special case c = 1 of Saalschiitz’s identity (5.g7), so we

could have saved a lot of work by going to hypergeometrics directly!

What about Problem 7? That extra-menacing sum gives us the formula

(

n+l,

m-n, 1,

tm+l,

tm+$,

=12

n’

which is the first case we’ve seen with three lower parameters. So it looks

new. But it really isn’t; the left-hand side can be replaced by

(

n, m-n-l,

-t

tm,

trn-;

1 -1,

using exercise 26, and Saalschiitz’s identity wins again.

Well, that’s another deflating experience, but it’s also another reason to

appreciate the power of hypergeometric methods.

The convolution identities in Table 202 do not have hypergeometric

equivalents, because their term ratios are rational functions of k only when

t is an integer. Equations (5.64) and (5.65) aren’t hypergeometric even when

t = 1. But we can take note of what (5.62) tells us when t has small integer

values:

(

,~;q-~~;Jl)

f-+,2")/("+nZn);

(

$r,

;r+;, fr+$,

-n,

-n-is,

-n-is-i

;r+;,

;

r+l,

-n--is,

-n-is+;,

-n-$.5+5

The first of these formulas gives the result of Problem 7 again, when the

quantities (r, s,n) are replaced respectively by

(1,2n

+ 1

m, -1

n).

Finally, the “unexpected” sum (5.20) gives us an unexpected

hypergeo-

metric identity that turns out to be quite instructive. Let’s look at it in slow

motion. First we convert to an infinite sum,

q32-k

= 2”

k$m

The term ratio from (2m

k)! 2k/m! (m

k)! is 2(k

m)/(k

2m), so we

have a hypergeometric identity with

= 2:

(2mm)F(‘~~~l2)

= 22m,

integerm>O.

(5.98)

216 BINOMIAL COEFFICIENTS

But look at the lower parameter

‘-

2m’.

Negative integers are verboten, so

this identity is undefined!

It’s high time to look at such limiting cases carefully, as promised earlier,

because degenerate hypergeometrics can often be evaluated by approaching

them from nearby nondegenerate points. We must be careful when we do this,

because different results can be obtained if we take limits in different ways.

For example, here are two limits that turn out to be quite different when one

of the upper parameters is increased by

hFO

-lSE, -3

-2+e

a,,(l

+ (4;;k;i + (--1+4(4-3)(-2)

(--2+El(-l+EI2!

(-l+~l(~)(l+~l(

-3)1-2)(-l)

(-2+E)(-l+E)(E)3!

)

FzF(I:';zll)

lii(l+#$+O+O)

q+o+o

Similarly, we have defined (1;) = 0 = lime-c

(-2’)

;

this is not the same

lime.+7

(1;::)

= 1. The proper way to treat

(5.98)

as a limit is to realize

that the upper parameter -m is being used to make all terms of the series

tkaO

(2c:kk)2k zero for k

m; this means that we want to make the following

more precise statement:

(2mm)

liiF(y2;,“,12)

22m,

integerm>O.

(5.99)

Each term of this limit is well defined, because the denominator factor

(-2m)’

does not become zero until k. > 2m. Therefore this limit gives us exactly the

sum (5.20) we began with.

5.6 HYPERGEOMETRIC TRANSFORMATIONS

It should be clear by now that a database of known hypergeometric

closed forms is a useful tool for doing sums of binomial coefficients. We

simply convert any given sum into its canonical hypergeometric form, then

look it up in the table. If it’s there, fine, we’ve got the answer. If not, we can

add it to the database if the sum turns out to be expressible in closed form.

We might also include entries in the table that say, “This sum does not have a

simple closed form in general.” For example, the sum xkSrn

(L)

corresponds

5.6 HYPERGEOMETRIC TRANSFORMATIONS 217

to the hypergeometric

(~)(A2

l-1))

integers n 3 m 3 0;

The

hypergeo-

metric database

should really be a

“knowledge base.”

this has a simple closed form only if m is near 0,

in,

or n.

But there’s more to the story, since hypergeometric functions also obey

identities of their own. This means that every closed form for hypergeometrics

leads to additional closed forms and to additional entries in the database. For

example, the identities in exercises 25 and 26 tell us how to transform one

hypergeometric into two others with similar but different parameters. These

can in turn be transformed again.

In 1793, J. F. PfafI discovered a surprising reflection law,

&F(a’cbl+)

F(a’;-blz),

(5.101)

which is a transformation of another type. This is a formal identity in

power series, if the quantity

(-z)“/(

z)~+~

is replaced by the infinite series

(--z)k(l

+ (":")z+

(k+;+'

)

. .) when the left-hand side is expanded (see

exercise 50). We can use this law to derive new formulas from the identities

we already know, when

# 1.

For example, Kummer’s formula (5.94) can be combined with the reflec-

tion law (5.101) if we choose the parameters so that both identities apply:

k$$b-a)~,

(5.102)

We can now set a = -n and go back from this equation to a new identity in

binomial coefficients that we might need some day:

2-,,

(b/4!

(b+n)!

b! (b/2+n)!

’

integer n 3 0. (5.103)

For example, when n = 3 this identity says that

l-3-

4.5

4.5.6

2(4 + b) 4(4 + b) (5 +

8(4 + b)(5 + b)(6 + b)

(b+3)(b+2)(b+l)

(b+6)(b+4)(b+2)

218 BINOMIAL COEFFICIENTS

It’s almost unbelievable, but true, for all b. (Except when a factor in the

denominator vanishes.)

This is fun; let’s try again. Maybe we’ll find a formula that will really

astonish our friends. What

Idoes

Pfaff’s reflection law tell us if we apply it to

the strange form (s.gg), where

= 2? In this case we set a =

-m,

b = 1,

and c =

-2mf

obtaining

lim

(-m)“(-2m-

1 + e)”

E'O

k>O

(-2m

+ c)k

because none of the limiting terms is close to zero. This leads to another

miraculous formula,

(-2)k

(-,yy2,

-l/2

=l/(

’

integer m 3 0.

(5.104)

When m = 3, for example, the sum is

and

(-y2)

is indeed equal to

-&.

When we looked at our binomial coefficient identities and converted them

to hypergeometric form, we overlooked (5.19) because it was a relation be-

tween two sums instead of a closed form. But now we can regard (5.19) as

an identity between hypergeometric series. If we differentiate it n times with

respect to y and then replace k by m

k, we get

m+r n+k

k>O

m-n-k

)(

)

m-n-k

-r

nfk

m-n-k

n (-X)m-n-k(X +

y)k.

This yields the following hypergeometric transformation:

a, -n

(

(a-c:)“F

=--

(-cp

(

a, -n

integer

1 -n+a-c

)

‘-’

’

n>O

(5.105)

5.6 HYPERGEOMETRIC TRANSFORMATIONS 219

Notice that when

= 1 this reduces to Vandermonde’s convolution, (5.93).

Differentiation seems to be useful, if this example is any indication; we

also found it helpful in Chapter 2, when summing x + 2x2 + . . . + nxn. Let’s

see what happens when a general hypergeometric series is differentiated with

respect to

(al+l)i;.

. .

a,(a,+l)kzk

(b,+l)“...b

n n

+l)kk!

al . . . a,

..b.

(5.10’3)

The parameters move out and shift up.

How do you pro-

It’s also possible to use differentiation to tweak just one of the parameters

while holding the rest of them fixed. For this we use the operator

flounce

4 ?

(Dunno,

but

7j$

calls it ?artheta’.)

which acts on a function by differentiating it and then

operator gives

multiplying by z. This

which by itself isn’t too useful. But if we multiply

parameters, say al, and add 4F, we get

F by one of its upper

al(al+l)‘ak...akzk

’

by.J&,

k?O n .

alF

al+l,

a2,

. . . . a,

bl,

. . . . b,

Only one parameter has been shifted.

220 BINOMIAL COEFFICIENTS

A similar trick works with lower parameters, but in this case things shift

down instead of up:

(bl

a!.

. .

k>O

(b,

-l)i;bi...b;k!

We can now combine all these operations and make a mathematical “pun”

Ever hear the one

by expressing the same quantity in two different ways. Namely, we have

about the brothers

who named their

(9+a,)...(4+a,)F

q =

al...a,F

altl,

. . . .

a,+1

cattle ranch Focus,

because it’s where

bl,

. . . .

the

sons raise meat?

and

1). . . (4 +

-- l)F

(bl-l)...(bn-1)F

,,“I”“‘~+),

...I

where F = F(al , . . . , a,;

, . . .

, b,;z). And (5.106) tells us that the top line

is the derivative of the bottom line. Therefore the general hypergeometric

function F satisfies the differential equation

D(9 +

1). . .

b,,

l)F = (4 + al). . .

a,)F,

(5.107)

where D is the operator

This cries out for an example. Let’s find the differential equation satisfied

by the standard a-over-1 hypergeometric series F(z) = F(a, b; c;

z).

According

(5.107),

we have

D(9+c-1)F

(i?+a)(4+b)F.

What does this mean in ordinary

notation

Well, (4 + c

l)F is zF’(z) +

1 )F(z), and the derivative of this gives the left-hand side,

F’(z) + zF”(z) + (c

l)F’(z)

5.6 HYPERGEOMETRIC TRANSFORMATIONS 221

On the right-hand side we have

(B+a)(zF’(z)+bF(z))

zi(zF’(z)+bF(z))

a(tF’(z)+bF(z))

zF’(z)+z’F”(z)+bzF’(z)+azF’(z)+abF(z).

Equating the two sides tells us that

~(1

-z)F”(z)+

(c-z(a+b+l))F’(z)

-abF(z)

= 0.

(5.108)

This equation is equivalent to the factored form (5.107).

Conversely, we can go back from the differential equation to the power

series. Let’s assume that F(z) =

kaO

tkzk is a power series satisfying (5.107).

A straightforward calculation shows that we must have

tk+l

(k+al)...(k+a,)

(k+b,)...(k+b,)(k+l)’

hence F(z) must be

F(al, . . . , a,,,;

bl,.

. . ,

b,;

z). We’ve proved that the

hypergeometric series (5.76) is the only formal power series that satisfies the

differential equation (5.107) and has the constant term 1.

It would be nice if hypergeometrics solved all the world’s differential

equations, but they don’t quite. The right-hand side of (5.107) always expands

into a sum of terms of the form

c%kzkFiki

(z), where

Flk’(z)

is the kth derivative

DkF(k); the left-hand side always expands into a sum of terms of the form

fikzk

‘Fikl(z)

with k > 0. So the differential equation (5.107) always takes

the special form

z”-‘(p,,

-zc~,JF(‘~(z)

+ . . . + ((3,

za,)F’(z)

ocoF(z) = 0.

The

function

F(z) = (1

-2)’

satisfies

8F = ~(4

r)F.

This

nives

another

proofYof the bino-

mial theorem.

Equation (5.108) illustrates this in the case n = 2. Conversely, we will prove

in exercise 6.13 that any differential equation of this form can be factored in

terms of the 4 operator, to give an equation like (5.107). So these are the dif-

ferential equations whose solutions are power series with rational term ratios.

Multiplying both sides of (5.107) by

dispenses with the D operator and

gives us an instructive all-4 form,

4(4 +

1). . . (4 + b,

l)F = ~(8 + al). .

a,)F.

The first factor 4 = (4+ 1

1) on the left corresponds to the (k+ 1) in the term

ratio (5.81), which corresponds to the k! in the denominator of the kth term

in a general hypergeometric series. The other factors (4 +

1) correspond

to the denominator factor (k+ bi), which corresponds to b: in (5.76). On the

right, the

corresponds to zk, and (4 +

) corresponds to af.

222 BINOMIAL COEFFICIENTS

One use of this differential theory is to find and prove new transforma-

tions. For example, we can readily verify that both of the hypergeometrics

satisfy the differential equation

~(1

-z)F"(z)

(afb

;)(l

-2z)F'(z)

-4abF(z)

= 0;

hence Gauss’s identity [116, equation

1021

(5.110)

must be true. In particular,

ICaution:

can’t

use (5.110) safely

,:4;:;

1;)

F(o+4;IT-11’)

’

whenever both infinite sums converge.

(5.111)

when

Izl

l/Z,

unless

both sides

are polynomials;

see

exercise 53.)

Every new identity for hypergeometrics has consequences for binomial

coefficients, and this one is no exception. Let’s consider the sum

&(m,k)(m+r+l)

(q)“,

integersm>n>O.

The terms are nonzero for 0 < k < m

n, and with a little delicate limit-

taking as before we can express this sum as the hypergeometric

liio

(

n-m,

-n-m-lfae

-m+

The value of

doesn’t affect the limit, since the nonpositive upper parameter

m cuts the sum off early. We can set

= 2, so that (5.111) applies.

The limit can now be evaluated because the right-hand side is a special case

of (5.92). The result can be expressed in simplified form,

gm,k)(m+,+l)

(G)

= ((m+nn1’2)2nPm[m+n is even],

~~~~o,

(5.112)

as shown in exercise 54. For example, when m = 5 and n = 2 we get

(z)(i)

($($/2

(:)(;)/4

(z)(i)/8

= 10

24 + 21

= 0; when m = 4

and n = 2, both sides give z.

5.6 HYPERGEOMETRIC TRANSFORMATIONS 223

We can also find cases where (5.110) gives binomial sums when z = -1,

but these are really weird. If we set a =

and b =

-n,

we get the

monstrous formula

These hypergeometrics are nondegenerate polynomials when n $ 2 (mod 3);

and the parameters have been cleverly chosen so that the left-hand side can

be evaluated by

(5.94).

We are therefore led to a truly mind-boggling result,

integer n 3 0, n

(mod 3).

(5.113)

This is the most startling identity in binomial coefficients that we’ve seen.

Small cases of the identity aren’t even easy to check by hand. (It turns out

The only use of

(5.113)

is to demon-

that both sides do give

when n = 3.) But the identity is completely useless,

strate the existence

of course; surely it will never arise in a practical problem.

incredibly useless

So that’s our hype for hypergeometrics. We’ve seen that hypergeometric

identities.

series provide a high-level way to understand what’s going on in binomial

coefficient sums. A great deal of additional information can be found in the

classic book by Wilfred N. Bailey

[15]

and its sequel by Lucy Joan Slater

[269].

5.7

PARTIAL HYPERGEOMETRIC SUMS

Most of the sums we’ve evaluated in this chapter range over all in-

dices k 3 0, but sometimes we’ve been able to find a closed form that works

over a general range 0 6 k < m. For example, we know from (5.16) that

integer m.

(5.114)

The theory in Chapter 2 gives us a nice way to understand formulas like this:

If f(k) = Ag(k) = g(k + 1)

g(k), then we’ve agreed to write

f(k)

g(k) + C, and

xbf(k)6k

= g(k)

I”,

= g(b)

g(a).

Furthermore, when a and b are integers with a < b, we have

tbf(k)Bk

f(k) = g(b)-g(a).

a<k<b

224 BINOMIAL COEFFICIENTS

Therefore identity (5.114) corresponds to the indefinite summation formula

(-l)%k

(-l)k-’

and to the difference formula

A((-lik(;))

= (-l)k+l

(;I;).

It’s easy to start with a function g(k) and to compute Ag(k) = f(k), a

function whose sum will be g(k) + C. But it’s much harder to start with f(k)

and to figure out its indefinite sum

f(k)

= g(k) + C; this function g

might not have a simple form. For example, there is apparently no simple

form for

(E)

6k;

otherwise we could evaluate sums like xkSn,3

(z)

, about

which we’re clueless.

In 1977, R. W. Gosper

[124]

discovered a beautiful way to decide whether

a given function is indefinitely summable with respect to a general class of

functions called hypergeometric terms. Let us write

al,

b,,

. .

..b.,

z =

a, . . . a,

by.

. .

(5.115)

for the kth term of the hypergeometric series F( al,. . . , a,,,; bl , . . . ,

b,;

z). We

will regard F(

al,.

. . , a,; bl , . . .

, b,;

z)k

as a function of k, not of

Gosper’s

decision procedure allows us to decide if there exist parameters c,

Al,

. . . , AM,

BI, . . . .

BN,

and Z such that

al, . . . .

b,,

.,.,

AI,

. . . ,

BI,

. . . , BN

(5.4

given al, . . . , a,,

bl,

. . . , b,, and

We will say that a given function

F(al,.

. .

,am;b,,.

. .

, bn;z)k is summable in hypergeometric terms if such

constants C, Al, . . . ,

AM,

Bl, . . . ,

BN,

Z exist.

Let’s write t(k) and T(k) as abbreviations for

F(al

, . . . , a,,,;

bl,

. . . ,

b,;

z)k

and

F(A,,

. . . , AM;

B,,

. . .

BN;

Z)k,

respectively. The first step in Gosper’s

decision procedure is to express the term ratio

t(k+

(k+al)...(k+a,)z

t(k)

(k+b,)...(k+b,)(k+l)

in the special form

t(k+ 1)

p(k+

q(k)

-=-

p(k)

r(k+

(5.117)

5.7 PARTIAL HYPERGEOMETRIC SUMS 225

(Divisibility ofpoly-

nomials is analogous

to divisibility of

integers. For exam-

ple,

+ a)\q(kl

means that the quo-

tient q(k)/(k+

is a polynomial.

It’s well known

that

a)\q(k)

and only if

q(-or) = 0.)

where

q, and

are polynomials subject to the following condition:

a)\q(k)

and

B)\r(k)

==+

is not a positive integer.

(5.118)

This condition is easy to achieve: We start by provisionally setting p(k) =

q(k)=(k+a,)...(k+a,)z,andr(k)=(k+bl-l)...(k+b,-l)k;then

we check if (5.118) is violated. If q and

have factors (k + a) and (k +

(3)

where a

= N > 0, we divide them out of q and

and replace p(k) by

p(k)(k+oL-l)N-‘=

p(k)(k+a-l)(k+a-2)...(k+fi+l).

The new p, q, and

still satisfy

(5.117),

and we can repeat this process until

(5.118) holds.

Our goal is to find a hypergeometric term T(k) such that

t(k) = cT(k+ 1)

-CT(k)

(5.119)

for some constant c. Let’s write

CT(k) =

r(k) s(k) t(k)

p(k)

’

(5.120)

(Exercise 55 ex-

where s(k) is a secret function that must be discovered somehow. Plugging

plains why we might

want to make this

(

5.120) into (5.117) and (5.119) gives us the equation that s(k) must satisfy:

magic substitution.)

p(k) = q(k)s(k+ 1)

-r(k)s(k)

(5.121)

If we can find s(k) satisfying this recurrence, we’ve found

t(k) 6k.

We’re assuming that T(k+ 1 )/T(k) is a rational function of k. Therefore,

by (5.120) and (5.11g), r(k)s(k)/p(k) =

T(k)/(T(k

+ 1) -T(k)) is a rational

function of k, and s(k) itself must be a quotient of polynomials:

s(k)

f(k)/g(kl.

(5.122)

But in fact we can prove that s(k) is itself a polynomial. For if g(k) #

and if f(k) and g(k) have no common factors, let N be the largest integer

such that (k + 6) and (k +

+ N

1) both occur as factors of g(k) for some

complex number

The value of N is positive, since N = 1 always satisfies

this condition. Equation (5.121) can be rewritten

p(k)g(k+l)g(k)

q(k)f(k+l)g(k)

-r(k)g(k+l)f(k),

and if we set k =

and k =

-6

N we get

r(-B)g(l-B)f(-6)

q(-B-N)f(l-B-N)g(-B-N)

226 BINOMIAL COEFFICIENTS

Now

f(-b)

# 0 and

f(l

-N) # 0, because f and g have no common

roots. Also

g(1

l3)

# 0 and g(-(3

N) # 0, because g(k) would otherwise

contain the factor

(k+

or (k+

+N), contrary to the maximality of N.

Therefore

T--f')

q(-8-N)

But this contradicts condition (5.118). Hence s(k) must be a polynomial.

The remaining task is to decide whether there exists a polynomial s(k)

satisfying (5.121), when p(k), q(k), and r(k) are given polynomials. It’s easy

to decide this for polynomials of any particular degree d, since we can write

s(k) =

cXdkd

(xdp,

kdm~’

-1-

. +

olo

for unknown coefficients

(&d,

. . . ,

o(o)

and plug this expression into the defin-

ing equation. The polynomial s(k) will satisfy the recurrence if and only if

the a’s satisfy certain linear equations, because each power of k must have

the same coefficient on both sides of (5.121).

But how can we determine the degree of s? It turns out that there

actually are at most two possibilities. We can rewrite (5.121) in the form

&(k)

Q(k)(s(k+

+s(k))

R(k)(s(k+

1) -s(k)),

where Q(k) = q(k) -r(k) and R(k) = q(k)

+r(k).

(5.123)

If s(k) has degree d, then the sum s(k + 1)

s(k) = 2adkd + . . . also has

degree d, while the difference s(k + 1)

s(k) = As(k) = dadkd-’ + . . . has

degree d

1. (The zero polynomial can be assumed to have degree -1.) Let’s

write deg(p) for the degree of a polynomial p. If deg(Q) 3 deg(R), then

the degree of the right-hand side of (5.128) is deg(Q) + d, so we must have

d = deg(p)

deg(Q). On the other hand if deg(Q)

deg(R) = d’, we can

write Q(k) =

@kd’-’

. . and R(k) =

ykd’

. . where y # 0; the right-hand

side of (5.123) has the form

(2,-?%

+ yd ,d)kd+d’-’ + . . . .

Ergo, two possibilities: Either 28 + yd # 0, and d = deg(p)

deg(R) + 1;

or 28 + yd = 0, and d > deg(p)

deg(R) + 1. The second case needs to be

examined only if

-2B/y

is an integer d greater than deg(p)

deg(R) + 1.

Thus we have enough facts to decide if a suitable polynomial s(k) exists.

If so, we can plug it into (5.120) and we have our T. If not, we’ve proved that

t(k) 6k is not a hypergeometric term.

5.7 PARTIAL HYPERGEOMETRIC SUMS 227

Time for an example. Let’s try the partial sum (5.114); Gosper’s method

should be able to deduce the value of

for any fixed n. Ignoring factors that don’t involve k, we want the sum of

The first step is to put the term ratio into the required form (5.117); we have

t(k+

(k-n)

P(k+

q(k)

t(k)

(k+

p(k)r(k+

Why isn’t

r(k) = k + 1 ?

Oh,

see.

so we simply take p(k) =

q(k) = k

n, and r(k) = k. This choice of

and

satisfies (5.118), unless n is a negative integer; let’s suppose it

isn’t. According to

(5.1~3)~

we should consider the polynomials Q(k) = -n

and R(k) = 2k

n. Since R has larger degree than Q, we need to look at

two cases. Either d = deg(p)

deg(R) +

which is 0; or d =

-26/y

where

= -n and y = 2, hence d = n. The first case is nicer, so let’s try it first:

Equation (5.121) is

1 =

(k-n)cxc-k%

and so we choose

0~0

= -l/n. This satisfies the required conditions and gives

CT(k) =

r(k)

s(k)

t(k)

p(k)

-,(li

.-.

(-l)k

n-l

k-,

(-W’

which is the answer we were hoping to confirm.

If we apply the same method to find the indefinite sum

(z) 6k, without

the (-1

)k,

everything will be almost the same except that q(k) will be n

hence Q(k) = n

2k will have greater degree than R(k) = n, and we will

conclude that d has the impossible value

deg(p)’

deg(Q) = -1. Therefore

the function (c) is not summable in hypergeometric terms.

However, once we have eliminated the impossible, whatever

remains-

however improbable-must be the truth (according to S. Holmes

[70]).

When

we defined p, q, and

we decided to ignore the possibility that n might be a

228 BINOMIAL COEFFICIENTS

negative integer. What if it is? Let’s set n = -N, where N is positive. Then

the term ratio for

(z) 6k is

t(k+

1) -(k+N)

p&S

‘I q(k)

___

t(k)

(k+l) =

p(k)

r(k+

‘I

and it should be represented by p(k) = (k+

l)Npl,

q(k)

-1, r(k)

Gosper’s method now tells us to look for a polynomial s(k) of degree d = N -1;

maybe there’s hope after all. For example, when N = 2 we want to solve

k+ 1 =

-((k+

l)cxl

LXO)

(km, +

Equating coefficients of k and 1 tells us that

1 = -a1

oL1;

1 = -cc~-cx~-cQ;

hence s(k) =

-ik

is a solution, and

CT(k) =

l+;k-$(,2)

k+l

Can this be the desired sum? Yes, it checks out:

= (-l)k(k+l) = i2 .

(

We can write the summation formula in another form,

(-‘y-l

This representation conceals the fact that ( ,‘) is summable in hypergeometric

terms, because [m/21 is not a hypergeometric term.

A catalog of summable hypergeometric terms makes a useful addition

to the database of hypergeometric sums mentioned earlier in this chapter.

Let’s try to compile a list of the sums-in-hypergeometric-terms that we know.

The geometric series

6k is a very special case, which can be written

tzk6k=(z-l))‘zk+Cor

~F(l;‘+)*,k

-&F(l;‘l~k+C.

(5.124)

“Excellent,

Holmes!”

“Elementary, my

dear Wa hon. ”

5.7 PARTIAL HYPERGEOMETRIC SUMS 229

We also computed

kzk

in Chapter 2. This summand is zero when

k = 0, so we get a more suitable hypergeometric term by considering the sum

1 )zk 6k instead.

appropriate formula turns out to be

(5.125)

in hypergeometric notation.

There’s also the formula

(k)

(,:,),

equation (5.10); we write it

k+;+l)

= (“‘,;t’) ,

to avoid division by zero, and get

,‘6k

&F(n+;‘l(‘)k,

-1.

(5.126)

Identity

(5.9)

turns out to be equivalent to this, when we express it

hyperge-

ometrically.

In general if we have a summation formula of the form

al, . . . .

a,,

kbk =

AI,

. . . . AM,

. . . .

'5,

. . . ,

k’

(5.127)

then we also have

al,

. . . .

a,,

bl,

. . . .

k+l

’

for any integer

There’s a general formula for shifting the index by

al,

i i

bl,

. . . .

a, . . . a,

fl,

. . . ,

a,+4

k+l

. . .

bl+1,

. . . ,

b,+l,l+l

’

k’

Hence any given identity (5.127) has an infinite number of shifted forms:

+1, . . . ,

a,+4

bltl,

. . . .

b,+l

=c”

..bT,

Ai...AT,

A1+1,

. .

..AM+~.

a\ . . . a,

B:.

. . BL

Blfl,

. . . . BN+~

’ k’

(5.128)

There’s usually a fair amount of cancellation among the a’s, A’s, b’s, and

B’s here. For example, if we apply this shift formula to

(5.126),

we get the

general identity

k6k =

sF(n+;';'lll)k,

(5.129)

230 BINOMIAL COEFFICIENTS

valid for all n # -1. The shifted version of (5.125) is

-1 L+l/(l-2)

ZZ---

l-z

1+1

(5.130)

With a bit of patience, we can compute a few more indefinite summation

identities that are potentially useful:

2+(1-a)z/(l-z),

l+(l-a)z/(l-z),2

a, b,

c+l,

(c-ab)/(c-a-b+l),

c+l,

a+b-c+l

(c)(c-b-a)

a)(c

(,,“dI;l,j

‘)k.

(5.133)

Exercises

Warmups

What is 1 l4 ? Why is this number easy to compute, for a person who

knows binomial coefficients?

For which value(s) of k is

(i)

a maximum, when n is a given positive

integer? Prove your answer.

Prove the hexagon property,

(;I:)

(k:,)

(nk+‘)

= (“i’) (i,‘:) (,“,).

Evaluate (-,‘) by negating (actually un-negating) its upper index.

Let p be prime. Show that

(F)

mod p = 0 for 0 < k < p. What does this

imply about the binomial coefficients (“i’)?

Fix up the text’s derivation in Problem 6, Section 5.2, by correctly ap-

A caseof

plying symmetry.

mistaken identity.

Is (5.34) true also when k < O?

5 EXERCISES 231

Evaluate

(L)(-l)k(l

-k/n)“. What is the approximate value of this

sum, when n is very large? Hint: This sum is

f (0) for some function f.

Show that the generalized exponentials of (5.58) obey the law

&t(z)

&(tz)

1/t

if t # 0,

where

E(z)

is an abbreviation for

&I(Z).

10 Show that

-2(ln(l

-2)

+ z)/

is a hypergeometric function.

11 Express the two functions

25 2'

sin2 =

z--+--rlt

arcsinz

1.23

1.3.25 1.3.5.27=

2.4.5

+ 2.4.6.7

+"'

in terms of hypergeometric series.

Which of the following functions of k is a “hypergeometric term,” in the

sense of

(5.115)?

Explain why or why not.

nk.

kn.

(Here t and T

aren’t necessar-

ily related as in

~w9~J

(k! + (k+

1)!)/2.

Hk,

that is, f +

. . + t.

e t(k)T(n

k)/T(n), when t and T are hypergeometric terms.

f (t(k) + T(k))/2, when t and T are hypergeometric terms.

(at(k) + bt(k+l) +

ct(k+2))/(a

+ bt(1) + ct(2)), when t is a

hypergeometric term.

Basics

13 Find relations between the superfactorial function P, =

nl,

k! of ex-

ercise 4.55, the hyperfactorial function

Q,,

= nL=,

kk,

and the product

I-I;==,

(;>.

Prove identity (5.25) by negating the upper index in Vandermonde’s con-

volution (5.22). Then show that another negation yields (5.26).

15 What is

(L)"(-l)"? Hint: See (5.29).

16 Evaluate the sum

(o:Uk)

(b:bk)

(c:k)(-li*

when a, b, c are nonnegative integers.

17 Find a simple relation between (2n;“2) and (2n;i’2).

232 BINOMIAL COEFFICIENTS

18 Find an alternative form analogous to (5.35) for the product

(;)

(r-y)

(r-y).

Show that the generalized binomials of (5.58) obey the law

2&(z)

tBp,(-z)-‘.

20 Define a “generalized bloopergeometric series” by the formula

al,

. . . ,

a!.

. ,

bl,

. . . . b,

k>O

b+...b$

k!’

using falling powers

inst,ead

of the rising ones in (5.76). Explain how

related to F.

21 Show that Euler’s definition of factorials is consistent with the ordinary

definition, by showing that the limit in (5.83) is

((m

1) . . . (1)) when

= m is a positive integer.

22 Use (5.83) to prove the factorial duplication formula:

x! (x

i)!

= (2x)!

(-;)!/22”.

23 What is the value of

F(-n,

;

1 )?

24 Find

(,,,tk) (“$“)4” by using hypergeometric series.

25 Show that

(a1

bl)

al,

a2,

bl+1,

bz, . . . . b,

= alF

al+l,

al,

. . . . a,

bl+l,

b2, . . . . b,

--b,F(“d~:~~::;~:“bniL).

Find a similar relation between the hypergeometrics F( al, al,

. . . , a,;

bl,...

,bn;z), F(al +

‘l,az,as

. . . .

a,;bl,...,

b,;z),

and F(al,az + 1,

as.. . , a,;

bl,.

. . ,

b,;z).

26 Express the function G(z) in the formula

al, . . . .

bl, . . . . b,

z = 1 + G(z)

as a multiple of a hypergeometric series.

By the way,

(-i)!

fi.

5 EXERCISES 233

27 Prove that

al,

al+;,

. . . . a,,

a,+;

b,,b,+;

,...,

b,,b,+;,;

(2m-n-1

z)2

2a1,...,2am

2b1,...,2b,

28 Prove Euler’s identity

= (,

+-a-bF

(c-a;-blg

by applying

Pfaff’s

reflection law (5.101) twice.

29 Show that confluent hypergeometrics satisfy

e’F(;i-z)

F(b;aiz).

30 What hypergeometric series F satisfies zF’(z) + F(z) =

l/(1

z)?

31 Show that if f(k) is any function summable in hypergeometric terms,

then f itself is a multiple of a hypergeometric term. In other words, if

f(k) 6k = cF(A,, . . .

,AM;

Bl,.

. . ,

BN;

Z)k

+ C, then there exist con-

stants

al,

. . . , a,,

bl,

. . . , b,, and z such that f(k) is a constant times

al,

. . . , a,;

bl , . . . ,

b,;

z)k.

Find

by Gosper’s method.

33 Use Gosper’s method to find

6k/(k2

1).

34 Show that a partial hypergeometric sum can

always

be represented as a

limit of ordinary hypergeometrics:

= F.o

E-C,

bl,

. . . ,

when c is a nonnegative integer. Use this idea to evaluate xkbm

(E)

(-1

)k.

Homework exercises

35 The notation tkG,, (;)2”-” is ambiguous without context. Evaluate it

as a sum on k;

as a sum on n.

Let

be the largest power of the prime p that divides

(“‘z”),

when m

and n are nonnegative integers. Prove that k is the number of carries

that occur when m is added to n in the radix p number system. Hint:

Exercise 4.24 helps here.

234 BINOMIAL COEFFICIENTS

Show that an analog of the binomial theorem holds for factorial powers.

That is, prove the identities

for all nonnegative integers n.

Show that all nonnegative integers n can be represented uniquely in the

formn =

(y)+(:)+(i)

ere a, b, and c are integers with 0 6 a < b < c.

(This is called the binomial number system.)

Show that if xy = ax -t by then xnyn =

xE=:=,

(‘“;~,~“)

(anbnpkxk +

an-

kbnyk) for all n > 0. Find a similar formula for the more general

product xmyn.

Find a closed form for

integers

m,n

3 0.

Evaluate

(L)k!/(n

+ 1 + k)!

when n is a nonnegative integer.

Find the indefinite sum

(( -1

)“/(t))

6x, and use it to compute the sum

xL=,(-l)“/(L)

in closed form.

Prove the triple-binomial identity (5.28). Hint: First replace

(iz:)

(m&-j>

(!I’

Use identity (5.32) to find closed forms for the double sums

~(-l)“k(i~k)

(3)

(L)

(m’~~~-k)

and

jF,ll)j+k(;)

(l;)

(bk)

(:)/(;x)

’

given integers m 3 a 3 0 and n 3 b 3 0.

Find a closed form for

tks,,

(234-k.

Evaluate the following

s’um

in closed form, when n is a positive integer:

Hint: Generating functions win again.

5 EXERCISES 235

47 The sum

(rkk+s)

(‘“;~~~“)

is a polynomial in

and s. Show that it

doesn’t depend on s.

48 The identity

xkGn

(“Lk)2pk = 2n can be combined with

tk30

(“lk)zk

l/(1

n+’

to yield

tk>n

(“~“)2~” =2”. What is the hypergeometric

form of the latter identity?

49 Use the hypergeometric method to evaluate

50 Prove Pfaff’s reflection law (5.101) by comparing the coefficients of 2” on

both sides of the equation.

51 The derivation of (5.104) shows that

lime+0 F(-m, -2m

1 +

-2m +

(-z2)

In this exercise we will see that slightly different limiting processes lead

to distinctly different answers for the degenerate hypergeometric series

-m,

-2m

1; -2m; 2).

Show that lim

e+~

F(-m +

-2m

1; -2m + 2e; 2) = 0, by using

Pfaff’s reflection law to prove the identity F(a, -2m

1; 2a; 2) = 0

for all integers m 3 0.

b What is lim

e+~

F(-m +

-2m

1; -2m +

2)?

52 Prove that if N is a nonnegative integer,

br].

= a, . . .

l-bl-N,..

. ,

l-b,-N,-N

1-al-N,...,l-am--N

53 If we put b =

-5

and z = 1 in Gauss’s identity (5.110), the left side

reduces to -1 while the right side is

fl.

Why doesn’t this prove that

-1

=+l?

54 Explain how the right-hand side of (5.112) was obtained.

55 If the hypergeometric terms t(k) = F(al , . . . , a,,,;

bl,

. . , ,

b,;

z)k and

T(k) =

F(A,,...

,AM;B~,...,BN;Z)~ satisfy t(k) = c(T(k+ 1) -T(k))

for all k 3 0, show that

= Z and m

n = M

56 Find a general formula for

(i3) 6k using Gosper’s method. Show that

(-l)k-’

[y]

is also a solution.

236 BINOMIAL COEFFICIENTS

57 Use Gosper’s method to find a constant

such that

is summable in hypergeometric terms.

If m and n are integers with 0 6 m 6 n, let

m,n

Find a relation between

T,,,n

and T,-1 ,+I, then solve your recurrence

by applying a summation factor.

Exam problems

Find a closed form for

when m and n are positive integers.

Use Stirling’s approximation

(4.23)

to estimate (“,‘“) when m and n are

both large. What does your formula reduce to when m = n?

Prove that when p is prime, we have

for all nonnegative integers m and n.

Assuming that p is prime and that m and n are positive integers, deter-

mine the value of

(,‘$‘)

mod

p2.

Hint: You may wish to use the following

generalization of Vandermonde’s convolution:

k+k&+k

JI:)(~)-~(~)

(r’+r2+i-~+Tm)*

Find a closed form for

given an integer n

5 EXERCISES 237

given an integer n 3 0.

65 Prove that

Ck(k+l)!

= n.

66 Evaluate “Harry’s double sum,’

as a function of m. (The sum is over both j and k.)

67 Find a closed form for

($))

(‘“nik)

integer n 3 0.

68 Find a closed form for

integer n > 0.

69 Find a closed form for69 Find a closed form for

min

kl ,...,k,>O

k,+...+k,=n

as a function of m and n.

70 Find a closed form for

ri)

(+)”

, integer n 3 0.

71 Let

where m and n are nonnegative integers, and let A(z) =

tk,o

okzk

the generating function for the sequence

(CQ,

al,

al,.

. .

a Express the generating function

S(z)

S,Z” in terms of A(z).

b Use this technique to solve Problem 7 in Section 5.2.

238 BINOMIAL COEFFICIENTS

Prove that, if m, n, and k are integers and n > 0,

n2k-v(k)

is an integer,

where v(k) is the number of l’s in the binary representation of k.

Use the repertoire method to solve the recurrence

= a;

(n-1)(X,-j

+X,-2),

for n

Hint: Both n! and ni satisfy this recurrence.

74 This problem concerns a deviant version of Pascal’s triangle in which the

sides consist of the numbers

2, 3, 4, . . . instead of all l’s, although the

interior numbers still satisfy the addition formula:

2 2

343 :

S’

4 7 7 4

ii0

i/.,

((t))

denotes the kth number in row n, for 1 < k < n, we have

((T))

((t))

= n, and

((L))

= ((“,‘)) + ((:I:)) for 1 < k < n. Express

the quantity

((i))

in closed form.

Find a relation between the functions

”

(n)

;

(31;:

’

S2(n)

(Sk”,

and the quantities

12”/.3J

and

[2n/31.

76 Solve the following recurrence for

n,O

Qo,k

= [k=Ol;

n,k

= Qn-l,k + Qn-l,k-, +

for n, k

5 EXERCISES 239

77 What is the value of

O<k

<,,

(kc’)

’

ifm>l?

.II!rn\

78 Assuming that m is a positive integer, find a closed form for

kmodm

(2kf

1) mod

(2m+

79 a What is the greatest common divisor of (:“),

(‘3”)

, . . . , (2tT,)? Hint:

Consider the sum of these n numbers.

b Show that the least common multiple of

(i)

(y)

, . . . , (E) is equal

to L(n +

l)/(n

+ 1), where L(n) =

lcm(l,2,.

. .

,n).

80 Prove that (L) < (en/k)k for all integers

k,n

3 0.

81 If 0 <

< 1 and 0 6 x 6

and if

m, n are nonnegative integers with

m < n, prove the inequality

(wm~'~(;)(~~;)xk

> 0.

Hint: Consider taking the derivative with respect to x.

Bonus problems

82 Prove that Pascal’s triangle has an even more surprising hexagon prop-

erty than the one cited in the text:

@((;I:),

(kg,)’

(n:l,)

gcd((“,‘),

(;+‘;),

(k”,))

if 0 < k < n. For example, gcd(56,36,210) = gcd(28,120,126) = 2.

83 Prove the amazing identity (5.32) by first showing that it’s true whenever

the right-hand side is zero.

84 Show that the second pair of convolution formulas, (5.61), follows from

the first pair, (5.60). Hint: Differentiate with respect to z.

85 Prove that

~il,m

m=l

l<kl<kz<...<k,,,$n

(k:+k:+.;+kL+Z”)

(-l)nn!3

’

(The left side is a sum of 2”

1 terms.) Hint: Much more is true

240 BINOMIAL COEFFICIENTS

86 Let al, . . . , a,, be nonnegative integers, and let C(al,. . . , a,,) be the

coefficient of the constant term

2:.

. .zt when the n(n

1) factors

are fully expanded into positive and negative powers of the complex vari-

ables

~1,

. . . . z,,.

Prove that

C(al

, . . .

, a,) equals the left-hand side of (5.31).

b Prove that if

21,

. . ,

z,,

are distinct complex numbers, then the

polynomial

f(4

k=l

l<j<n

j#k

is identically equal to 1.

Multiply the original product of n(n

1) factors by f (0) and deduce

that

C(al,al,...,a,)

isequalto

C(al

-l,az,...,

a,)+C(al,a2-l,...,a,)

+ . . .

+C(al,a2

,...,

a,-1).

(This recurrence defines multinomial coefficients, so C(al , . . . , a,)

must equal the right-hand side of

(5.31).)

8’7

Let m be a positive integer and let

= eni”“. Show that

rg-,(zm)n+’

= (1 +

m)K,(zm)

-m

(C2i+1zIBl+,,,(~2i+l~~l/m)n+1

osj<mTm+

1)%+l,,(L2j+l~)-l

(This reduces to (5.74) in the special case m =

1.)

88 Prove that the coefficients

in (5.47) are eqUa1 to

for all k > 1; hence

/ski

l/(k-

1).

5 EXERCISES 241

89 Prove that

(5.19)

has an infinite counterpart,

(mlr)Xk?Jm-k =

(ir)

(-X)k(X+y)“pk, integer m,

k>m

1x1

Iy/

and

Ix/

y/.

Differentiate this identity n times with

respect to y and express it in terms of hypergeometrics; what relation do

you get?

Problem 1 in Section 5.2 considers

tkaO

/(l)

when

and s are integers

with s 3 r 3 0. What is the value of this sum if r and s aren’t integers?

91 Prove Whipple’s identity,

ia,

;a+;,

l-a-b-c

l+a-b,

l+a-c

= (1

-z)“F

by showing that both sides satisfy the same differential equation.

92 Prove Clausen’s product identities

:+a,

$+b

1 +a+b

What identities result

formulas are equated?

=F(

$+a-b,

i--a+b

l+a+b,

l-a-b

when the coefficients of

2”

on both sides of these

93 Show that the indefinite sum

f(i)+a)

has a (fairly) simple form, given any function f and any constant a.

94 Show that if w =

e2ni/3

we have

k+l&x3n

(k,~m)2WL+2m

(n,;In)

’

integer

’

”

242 BINOMIAL COEFFICIENTS

Research problems

95 Let q(n) be the smallest odd prime factor of the middle binomial co-

efficient

(t).

According to exercise 36, the odd primes p that do not

divide (‘z) are those for which all digits in n’s radix p representation are

1)/2 or less. Computer experiments have shown that q(n) 6 11 for

all n <

101oooo,

except that q(3160) = 13.

Isq(n)<ll

foralln>3160?

b Is q(n) = 11 for infinitely many n?

A reward of

$(:)

(“,)

(z)

is offered for a solution to either (a) or (b).

96 Is

(‘,“)

divisible by the square of a prime, for all n > 4?

97 For what values of n is

(F)

E (-1)” (mod

(2n-t

l))?

Special Numbers

SOME SEQUENCES of numbers arise so often in mathematics that we rec-

ognize them instantly and give them special names. For example, everybody

who learns arithmetic knows the sequence of square numbers

(1,4,9,16,

. . ).

In Chapter 1 we encountered the triangular numbers

(1,3,6,10,

. . . ); in Chap-

ter 4 we studied the prime numbers

(2,3,5,7,.

.);

in Chapter 5 we looked

briefly at the Catalan numbers

(1,2,5,14,

. . .

In the present chapter we’ll get to know a few other important sequences.

First on our agenda will be the Stirling numbers {t} and [L] , and the Eulerian

numbers

(i);

these form triangular patterns of coefficients analogous to the

binomial coefficients

(i)

in Pascal’s triangle. Then we’ll take a good look

at the harmonic numbers H,, and the Bernoulli numbers

B,;

these differ

from the other sequences we’ve been studying because they’re fractions, not

integers. Finally, we’ll examine the fascinating Fibonacci numbers F, and

some of their important generalizations.

6.1 STIRLING NUMBERS

We begin with some close relatives of the binomial coefficients, the

Stirling numbers, named after James Stirling (1692-1770). These numbers

come in two flavors, traditionally called by the no-frills names “Stirling num-

bers of the first and second kind!’ Although they have a venerable history

and numerous applications, they still lack a standard notation. We will write

{t} for Stirling numbers of the second kind and [z] for Stirling numbers of

the first kind, because these symbols turn out to be more user-friendly than

the many other notations that people have tried.

Tables 244 and 245 show what

{f;}

and [L] look like when n and k are

small. A problem that involves the numbers “1, 7, 6, 1” is likely to be related

to {E}, and a problem that involves

“6,

11, 6, 1” is likely to be related to

[;I,

just as we assume that a problem involving “1, 4, 6, 4, 1” is likely to be

related to (c); these are the trademark sequences that appear when n = 4.

243

244 SPECIAL NUMBERS

Table 244 Stirling’s triangle for subsets.

q---mnni;)

(751

{aI

0 1

7 6

0 1 15 25 10

0 1 31 90 65 15

0 1 63

301

350

140 21 1

0 1

127 966

1701

1050

266 28 1

0 1 255 3025 7770 6951

2646 462 36

Stirling numbers of the second kind show up more often than those of

the other variety, so let’s consider last things first. The symbol {i} stands for

(Stirling

himself

the number of ways to partition a set of n things into k nonempty subsets.

For example, there are seven ways to split a four-element set into two parts:

~~~fi~d~~!

book

[281].)

{1,2,3IuI41,

u,2,4u31,

U,3,4IuI21,

12,3,4uUl,

{1,2IuI3,41,

Il,3ICJ{2,41,

u,4wv,3h

(6.1)

thus {i} = 7. Notice that curly braces are used to denote sets as well as

the numbers {t} . This notational kinship helps us remember the meaning of

which can be read “n subset

k!’

Let’s look at small k. There’s just one way to put n elements into a single

nonempty set; hence

{

‘,‘}

= 1, for all n > 0. On the other hand

{y}

= 0,

because a O-element set is empty.

The case k = 0 is a bit tricky. Things work out best if we agree that

there’s just one way to partition an empty set into zero nonempty parts; hence

{i} = 1. But a nonempty set needs at least one part, so {i} = 0 for n > 0.

What happens when k

Certainly {i} = 0. If a set of n > 0 objects

is divided into two nonempty parts, one of those parts contains the last object

and some subset of the first

objects. There are

2+’

ways to choose the

latter subset, since each of the first n

1 objects is either in it or out of it;

but we mustn’t put all of those objects in it, because we want to end up with

two

nonempty

parts. Therefore we subtract

T-1

)

integer n > 0.

(6.2)

(This tallies with our enumeration of {i} = 7 =

1 ways above.)

6.1 STIRLING NUMBERS 245

Table 245 Stirling’s triangle for cycles.

120

720

5040

40320

35 10

274

225 85 15 1

1764 1624

735

175 21 1

13068 13132 6769 1960

322 28 1

109584 118124 67284 22449 4536 546 36 1

A modification of this argument leads to a recurrence by which we can

compute {L} for all k: Given a set of n > 0 objects to be partitioned into k

nonempty parts, we either put the last object into a class by itself (in

{:I:}

ways), or we put it together with some nonempty subset of the first n

objects. There are

k{n,‘}

possibilities in the latter case, because each of the

{

“;‘}

ways to distribute the first n

1 objects into k nonempty parts gives

k subsets that the nth object can join. Hence

{;1)

k{rrk’}+{EI:},

integern>O.

This is the law that generates Table 244; without the factor of k it would

reduce to the addition formula (5.8) that generates Pascal’s triangle.

And now, Stirling numbers of the first kind. These are somewhat like

the others, but

[L]

counts the number of ways to arrange n objects into k

cycles instead of subsets. We verbalize ‘[;I’ by saying “n cycle k!’

Cycles are cyclic arrangements, like the necklaces we considered in Chap-

ter 4. The cycle

can be written more compactly as ‘[A, B, C, D]‘, with the understanding that

[A,B,C,D] = [B,C,D,A] = [C,D,A,Bl =

[D,A,B,Cl;

a cycle “wraps around” because its end is joined to its beginning. On the other

hand, the cycle [A, B, C,

is not the same as [A, B, D,

or [D, C, B, A].

246 SPECIAL NUMBERS

There are eleven different ways to make two cycles from four elements:

“There are nine

and sixty ways

[1,2,31

[41,

[’

,a41

[1,3,41

[&3,4 [II,

of constructing

[1,3,21

[41,

[’

,4,21

P,4,31

PI,

tribal lays,

And-every-single-

P,21

[3,41,

[’

,31

4 ,

[I,41

P,31;

one-of-them-is-

rjght,”

hence

[“;I

= 11.

-Rudyard Kipling

A singleton cycle (that is, a cycle with only one element) is essentially

the same as a singleton set (a set with only one element). Similarly, a 2-cycle

is like a 2-set, because we have [A,

[B,

A] just as {A, B} = {B, A}. But

there are two diflerent 3-cycles, [A, B,

and [A, C,

B].

Notice, for example,

that the eleven cycle pairs in (6.4) can be obtained from the seven set pairs

in (6.1) by making two cycles from each of the 3-element sets.

In general,

n!/n

= (n -- 1) ! cycles can be made from any n-element set,

whenever n > 0. (There are n! permutations, and each cycle corresponds

to n of them because any one of its elements can be listed first.) Therefore

we have

= (n-l)!, integer n > 0.

This is much larger than the value {;} = 1 we had for Stirling subset numbers.

In fact, it is easy to see that the cycle numbers must be at least as large as

the subset numbers,

[E]

{L}y

integers n, k 3 0,

because every partition into nonempty subsets leads to at least one arrange-

ment of cycles.

Equality holds in (6.6) when all the cycles are necessarily singletons or

doubletons, because cycles are equivalent to subsets in such cases. This hap-

pens when k = n and when k = n

1; hence

[Z] =

{iI}’

[nl:l]

{nil}

In fact, it is easy to see that.

[“n]

{II}

”

[nil]

{nnl}

= (I)

(6.7)

(The number of ways to arrange n objects into n

1 cycles or subsets is

the number of ways to choose the two objects that will be in the same cycle

or subset.) The triangular numbers (;) = 1, 3, 6, 10, . . . are conspicuously

present in both Table 244 and Table 245.

6.1 STIRLING NUMBERS 247

We can derive a recurrence for

[z]

by modifying the argument we used

for {L}. Every arrangement of n objects in k cycles either puts the last object

into a cycle by itself (in

[:::I

or inserts that object into one of the [“;‘Iy )

cycle arrangements of the first n- 1 objects. In the latter case, there are n- 1

different ways to do the insertion. (This takes some thought, but it’s not hard

to verify that there are j ways to put a new element into a j-cycle in order to

make a (j + 1)-cycle. When j = 3, for example, the cycle [A, B,

leads to

[A,

[A,B,D,Cl,

[A,D,B,Cl

when we insert a new element D, and there are no other possibilities. Sum-

ming over all j gives a total of n- 1 ways to insert an nth object into a cycle

decomposition of n

1 objects.) The desired recurrence is therefore

(n-l)[ni’]

[:I:],

integern>O.

This is the addition-formula analog that generates Table 245.

Comparison of (6.8) and (6.3) shows that the first term on the right side is

multiplied by its upper index (n- 1) in the case of Stirling cycle numbers, but

by its lower index k in the case of Stirling subset numbers. We can therefore

perform “absorption” in terms like n[z] and k{

T},

when we do proofs by

mathematical induction.

Every permutation is equivalent to a set of cycles. For example, consider

the permutation that takes 123456789 into 384729156. We can conveniently

represent it in two rows,

123456789

384729156,

showing that 1 goes to 3 and 2 goes to 8, etc. The cycle structure comes

about because 1 goes to 3, which goes to 4, which goes to 7, which goes back

to 1; that’s the cycle

[1,3,4,7].

Another cycle in this permutation is [2,8,5];

still another is

[6,91.

Therefore the permutation 384729156 is equivalent to

the cycle arrangement

[1,3,4,7l

L&8,51

691.

If we have any permutation

rr1

rrz

. . . rr, of

{

1,2,.

. . , n}, every element is in a

unique cycle. For if we start with mu = m and look at

rrmor

= rrm,,

etc., we must eventually come back to mk =

TQ.

(The numbers must re-

peat sooner or later, and the first number to reappear must be mc because

we know the unique predecessors of the other numbers ml,

ml,

. . . ,

m-1

Therefore every permutation defines a cycle arrangement. Conversely, every

248 SPECIAL NUMBERS

cycle arrangement obviously defines a permutation if we reverse the construc-

tion, and this one-to-one correspondence shows that permutations and cycle

arrangements are essentially the same thing.

Therefore

[L]

is the number of permutations of n objects that contain

exactly k cycles. If we sum

[z]

over all k, we must get the total number of

permutations:

= n!, integer n 3 0.

(6.9)

For example, 6 + 11 + 6 + 1 = 24 = 4!.

Stirling numbers are useful because the recurrence relations (6.3) and

(6.8) arise in a variety of problems. For example, if we want to represent

ordinary powers

x”

by falling powers xc, we find that the first few cases are

x0;

x1;

x2.+&

x3+3&+,1;

x4+6x3+7xL+x1,

These coefficients look suspiciously like the numbers in Table 244, reflected

between left and right; therefore we can be pretty confident that the general

formula is

Xk,

integer n 3 0.

(6.10)

We’d better define

{C}

[;I

when k < 0 and

And sure enough, a simple proof by induction clinches the argument: We n 3

have

= xk+l + kxk,

bec:ause

xk+l =

;

hence

xnP1

x${~;‘}x”

;,i”;‘}x”+;{“;‘}kx”

;,{;I;}x”‘Fj”;‘}kx”

;,(k{“;‘}

{;;;;})xh

{;}xh.

In other words, Stirling subset numbers are the coefficients of factorial powers

that yield ordinary powers.

6.1 STIRLING NUMBERS 249

We can go the other way too, because Stirling cycle numbers are the

coefficients of ordinary powers that yield factorial powers:

xiT

= xo.

Xi =

xiI

= x2 + x’ ;

x”

+3x2 +2x';

+6x3 +11x* +6x'.

We have (x+n-

l).xk

=xk+’

+ (n

1 )xk, so a proof like the one just given

shows that

(xfn-1)~~~’

(x+n-1);

ril]xk

[r;]xk.

This leads to a proof by induction of the general formula

integer n 3 0.

(6.11)

(Setting x = 1 gives (6.9) again.)

But wait, you say. This equation involves rising factorial powers

xK,

while

(6.10) involves falling factorials xc. What if we want to express

in terms of

ordinary powers, or if we want to express

X”

in terms of rising powers? Easy;

we just throw in some minus signs and get

integer n > 0; (6.12)

(6.13)

This works because, for example, the formula

x(x-1)(x-2)(x-3)

= x4-6x3+11x2-6x

is just like the formula

XT =

~(~+1)(~+2)(x+3)

= x4+6x3+11x2+6x

but with alternating signs. The general identity

(-ly-#

(6.14)

of exercise 2.17 converts (6.10) to (6.12) and (6.11) to (6.13) if we negate x.

250 SPECIAL NUMBERS

Table 250 Basic Stirling number identities, for integer n > 0.

Recurrences:

{L}

kjnk’}+{;I:}.

(n-

Special values:

{I}

[i]

q [n =

[n>Ol;

(2np'

-1)

[n>O];

{nnl}

[n”l]

=’

(1)’

{;} = [j =

(;)

= 1.

{;} =

[;I

(n-l)![n>O].

(n-l)!H,-1

[n>O]

if k > n.

Converting between powers:

=T-

Inversion formulas:

(-l)“pk

[m=n];

6.1 STIRLING NUMBERS 251

Table 251 Additional Stirling number identities, for integers 1, m, n 3 0.

{Z}

$(i){k}.

[zl]

[J(k).

{;}

;

(;){;‘t:J(w”.

[;I

[;I;]

($J-k.

m!{z}

(3kn(-1)--k.

{:I:}

&{L)(-+llnek.

[;;:I

f.[#~

&[j/k!.

{m+;+‘}

k{n;k}.

[m+:“]

g(n+k)[nlk].

(;)

F(nk++:}[$-li'"*.

In-m)!(Jh3ml

[;+':]{k}(-l)mek.

{n:m}

(ZZ)

(:I;)

[“:“I

[n:m]

~(Z~L)(ZI;)(-:“}

{lJm}(L:m)

G{F}{“m”)(L)

(6.15)

(6.16)

(6.17)

(6.18)

(6.19)

(6.20)

(6.21)

(6.22)

(6.23)

(6.24)

(6.25)

(6.26)

(6.27)

(6.28)

(6.29)

252 SPECIAL NUMBERS

We can remember when to stick the

(-l)“pk

factor into a formula like

(6.12) because there’s a natural ordering of powers when x is large:

Xii >

x5,

for all x > n > 1.

(6.30)

The Stirling numbers

[t]

and {z} are nonnegative, so we have to use minus

signs when expanding a “small” power in terms of “large” ones.

We can plug (6.11) into (6.12) and get a double sum:

This holds for all x, so the coefficients of x0, x1, . . . , xnp’, x”+‘, xn+‘, . . on

the right must all be zero and we must have the identity

;

(-l)“pk

[m=n],

integers

m,n

3 0.

Stirling numbers, like b.inomial coefficients, satisfy many surprising iden-

tities. But these identities aren’t as versatile as the ones we had in Chapter 5,

so they aren’t applied nearly as often. Therefore it’s best for us just to list

the simplest ones, for future reference when a tough Stirling nut needs to be

cracked. Tables 250 and 251 contain the formulas that are most frequently

useful; the principal identities we have already derived are repeated there.

When we studied binomial coefficients in Chapter 5, we found that it

was advantageous to define

1::)

for negative n in such a way that the identity

(;) =

(“,‘)

(;I:)

valid without any restrictions. Using that identity to

extend the (z)‘s beyond those with combinatorial significance, we discovered

(in Table 164) that Pascal’s triangle essentially reproduces itself in a rotated

form when we extend it upward. Let’s try the same thing with Stirling’s

triangles: What happens if we decide that the basic recurrences

{;}

k{n;‘}+{;I:}

(n-I)[“;‘]

[;I:]

are valid for all integers n and k? The solution becomes unique if we make

the reasonable additional stipulations that

{E}

= [k=Ol and

{t}

[z]

[n=O].

(6.32)

6.1 STIRLING NUMBERS 253

Table 253 Stirling’s triangles in tandem.

{:5}

{_nq}

{:3}

{:2}

{:1}

{i}

{Y}

{I}

{3}

{a}

{r}

-5

-4 10 1

-3 35 6 1

-2 50 11 3 1

-1 24 6 2 1

000001

0 0 0 0 0 0 1

0 0 0 0 0 0 11

0 0 0 0 0 0 13 1

0 0 0 0 0 0 17 6 1

0 0 0 0 0 0 115 25 10 1

In fact, a surprisingly pretty pattern emerges:

Stirling’s triangle for cycles

appears above Stirling’s triangle for subsets, and vice versa! The two kinds

of Stirling numbers are related by an extremely simple law:

[I]

{I:},

integers k,n.

We have “duality,” something like the relations between min and max, between

1x1

and

[xl,

between

and

xK,

between gcd and lcm. It’s easy to check that

both of the recurrences

= (n- 1) [“;‘I +

[i;:]

and

{i}

k{n;‘}

{:I:}

amount to the same thing, under this correspondence.

6.2

EULERIAN

NUMBERS

Another triangle of values pops up now and again, this one due to

Euler

[88,

page

4851,

and we denote its elements by (E). The angle brackets

in this case suggest “less than” and “greater than” signs; (E) is the number of

permutations

rr1

rr2

. . .

rr,

,2, . . . ,

n} that have k ascents, namely, k places

where

< nj+l. (Caution: This notation is even less standard than our

no-

tations

[t]

{i}

for Stirling numbers. But we’ll see that it makes good sense.)

For example, eleven permutations of

,2,3,4}

have two ascents:

1324, 1423, 2314, 2413, 3412;

1243, 1342, 2341;

2134, 3124, 4123.

(The first row lists the permutations with

7~2

7r3

7~;

the second row

lists those with

rrl

7~3

7~4

and

rr2

713

7r4.)

Hence

(42)

= 11.

254 SPECIAL NUMBERS

Table 254 Euler’s triangle.

11 11

66. 26

57 302 302 57

1 120

1191

2416

1191

120

1 247 4293 15619

15619 4293

247 1 0

1 502 14608 88234 156190 88234 14608

502 1 0

Table 254 lists the smallest Eulerian numbers; notice that the trademark

sequence is

11, 11, 1

this time. There can be at most n

ascents, when

n > 0, so we have

(:)

[n=:O]

on the diagonal of the triangle.

Euler’s triangle, like Pascal’s, is symmetric between left and right. But

in this case the symmetry law is slightly different:

(,-Y-k),

integer n> 0;

(6.34)

The permutation

rrr

7~2

. . .

71,

has n- 1 -k ascents if and only if its “reflection”

7rn

. .

7r27rl

has k ascents.

Let’s try to find a recurrence for

(i).

Each permutation p =

. . .

pnpl

of{l,...

leads to n permutations of

{1,2,.

. .

,n}

if we insert the new

element n in all possible ways. Suppose we put n in position j, obtaining the

permutation

. . .

pi-1

. . .

~~-1.

The number of ascents in

is the

same as the number in

if j = 1 or if

pi-1

pi;

it’s one greater than the

number in p, if

pi-1

or if j = n. Therefore

has k ascents in a total of

(kf

l)(n,‘)

rom permutations

that have k ascents, plus a total of

((n-2)-P-1)+1)(:X;)

ways from permutations

that have k- 1 ascents.

The desired recurrence is

[k+lJ(n,l>+[n-k](LI:>.

integern>O.

(6.35)

Once again we start the recurrence off by setting

= [k=O], integer k,

(6.36)

and we will assume that

(L)

= 0 when k < 0.

6.2 EULERIAN NUMBERS 255

Eulerian numbers are useful primarily because they provide an unusual

connection between ordinary powers and consecutive binomial coefficients:

F(L)(“L”>,

integern>O.

(This is “Worpitzky’s identity”

[308].)

For example, we have

(1)+(T))

(;)+qy)+(y’),

(;)+ll(x;')+11(Xfi2)+(X;3),

and so on.

It’s easy to prove (6.37) by induction (exercise 14).

Incidentally, (6.37) gives us yet another way to obtain the sum of the

first n squares: We have

($(i)

(f)

(“i’) = (i) + (ki’), hence

12+22+...+n2

((;)+(;)+-.+(;))+((;)+(;)+.-+(";'))

("p)

("f2)

;(

n+l)n((n-l)+(n+2)).

The Eulerian recurrence (6.35) is a bit more complicated than the Stirling

recurrences (6.3) and

(6.8),

so we don’t expect the numbers

(L)

to satisfy as

many simple identities. Still, there are a few:

(t)

(n:‘)(m+l

-k)“(-llk;

-!{Z}

G(E)(n*m)’

(;)

{;}(“,“)(-l)nPk-mk!

(6.38)

(6.39)

(6.40)

If we multiply

(6.39)

znPm

and sum on m, we get

x,,

{

t}m! zn-“’ =

(c)

+ 1)

Replacing

1 and equating coefficients of

gives

(6.40). Thus the last two of these identities are essentially equivalent. The

first identity,

(6.38),

gives us special values when m is small:

(i)

= 1;

(I)

2n-n-l;

(1)

3”-(n+l)Z”+(n:‘) .

256 SPECIAL NUMBERS

Table 256 Second-order Eulerian triangle.

{'l

L\‘,

10 1

J .f\

1 2

6 0

1 22

58 24

1 52

328 444 120

1 114

1452 4400 3708 720

1 240

5610 32120 58140

33984 5040

1 494

19950 195800 644020 785304 341136 40320

We needn’t dwell further on

Eulerian

numbers here; it’s usually sufficient

simply to know that they exist, and to have a list of basic identities to fall

back on when the need arises. However, before we leave this topic, we should

take note of yet another triangular pattern of coefficients, shown in Table 256.

We call these “second-order Eulerian numbers”

((F)),

because they satisfy a

recurrence similar to (6.35) but with n replaced by 2n

1 in one place:

((E))

(k+l)((n~1))+(2n-l-k)((~-:)>.

(6.41)

These numbers have a curious combinatorial interpretation, first noticed by

Gessel and Stanley

[118]:

If we form permutations of the multiset

(1,

1,2,2,

. ,n,n} with the special property that all numbers between the two occur-

rences of m are greater than m, for 1 6 m 6 n, then ((t)) is the number of

such permutations that have k ascents. For example, there are eight suitable

single-ascent permutations of

1,2,2,3,3}:

113322,

133221, 221331, 221133,

223311, 233211, 331122, 331221.

Thus

((T))

= 8. The multiset

{l,

1,2,2,.

. n}

, n,

has a total of

(2n-1)(2n-3)...(l)

(6.42)

suitable permutations, because the two appearances of n must be adjacent

and there are 2n

1 places to insert them within a permutation for n

For example, when n = 3 the permutation 1221 has five insertion points,

yielding

331221, 133221, 123321, 122331,

and

122133.

Recurrence (6.41) can

be proved by extending the argument we used for ordinary Eulerian numbers.

6.2

EULERIAN

NUMBERS 257

Second-order Eulerian numbers are important chiefly because of their

connection with Stirling numbers

[119]:

We have, by induction on n,

{x"n}

&(($~+n~lyk).

integern30;

(6.43)

[x”n]

g(~))(“~k)

For example,

{Zl}

(1))

integer n 3 0.

(6.44)

[x:1]

(1);

{z2}

(7’)

+2(g)

[x:2]

(:)+2(y);

(,r,}

(“:‘)

+8(y)

+6(d)

[xx3]

(1)

+8(x;‘)

+6(x12).

(We already encountered the case n = 1 in (6.7).) These identities hold

whenever x is an integer and n is a nonnegative integer. Since the right-hand

sides are polynomials in x, we can use (6.43) and (6.44) to define Stirling

numbers

{

.“,}

and [,Tn] for arbitrary real (or complex) values of x.

If n > 0, these polynomials

{

.“,} and

[,“J

are zero when x = 0, x =

. . . ,

and x = n; therefore they are divisible by (x-O), (x-l), . . . , and (x-n).

It’s interesting to look at what’s left after these known factors are divided out.

We define the Stirling polynomials o,(x) by the rule

&l(x)

[

.",

/(X(X-l)...(X-TX)).

(6.45)

(The degree of o,(x) is n

1.)

The first few cases are

l/x

isa

polynomial?

(Sorry

about

that.)

q)(x)

= l/x;

CT,(x)

l/2;

02(x)

(3x-1)/24;

q(x) =

(x2

-x)/48;

Q(X)

(15x3

-30x2+5x+2)/5760.

They can be computed via the second-order Eulerian numbers; for example,

CQ(X)

((~-4)(x-5)+8(x+1)(x-4)

+6(x+2)(x+1))/&

258 SPECIAL NUMBERS

Table 258 Stirline convolution formulas.

Ok(T)

0,-k(S) = (r + s)on(r +

k=O

(6.46)

k&(T) (T&(s) =

no,(r+

(6.47)

k=O

rS&(l.+k)On&S-4-k)

= (?'+S)D,(l-+S+n)

k=O

kCTk(T+k)G,~k(Si-

n-k) =

no,(r+S+n)

(6.48)

(6a)

k=O

(-l)""-l(mn!,),~"-,(m)

(6.50)

[

imT

k,(n)

(6.51)

It turns out that these polynomials satisfy two very pretty identities:

zez

’

(

)

XpJ,(X)2Q

TX>0

(iln&---

x&o,(x+n)zn;

(6.52)

(6.53)

Therefore we can obtain general convolution formulas for Stirling numbers, as

we did for binomial coefficients in Table 202; the results appear in Table 258.

When a sum of Stirling numbers doesn’t fit the identities of Table 250 or 251,

Table 258 may be just the ticket. (An example appears later in this chapter,

following equation (6.100).

Elxercise

7.19 discusses the general principles of

convolutions based on identit:ies like (6.52) and (6.53).)

6.3 HARMONIC NUMBERS

It’s time now to take a closer look at harmonic numbers, which we

first met back in Chapter 2:

,+;+;+...+;

f;,

integer n 3 0.

k=l

(6.54)

These numbers appear so often in the analysis of algorithms that computer

scientists need a special notation for them. We use

H,,

the ‘H’ standing for

6.3 HARMONIC NUMBERS 259

This must be

Table 259.

“harmonic,” since a tone of wavelength l/n is called the nth harmonic of a

tone whose wavelength is 1. The first few values look like this:

n101234

5 6 7 8 9 10

Exercise 21 shows that

is never an integer when n > 1.

Here’s a card trick, based on an idea by R. T. Sharp

[264],

that illustrates

how the harmonic numbers arise naturally in simple situations. Given n cards

and a table, we’d like to create the largest possible overhang by stacking the

cards up over the table’s edge, subject to the laws of gravity:

To define the problem a bit more, we require the edges of the cards to be

parallel to the edge of the table; otherwise we could increase the overhang by

rotating the cards so that their corners stick out a little farther. And to make

the answer simpler, we assume that each card is 2 units long.

With one card, we get maximum overhang when its center of gravity is

just above the edge of the table. The center of gravity is in the middle of the

card, so we can create half a cardlength, or 1 unit, of overhang.

With two cards, it’s not hard to convince ourselves that we get maximum

overhang when the center of gravity of the top card is just above the edge

of the second card, and the center of gravity of both cards combined is just

above the edge of the table. The joint center of gravity of two cards will be

in the middle of their common part, so we are able to achieve an additional

half unit of overhang.

This pattern suggests a general method, where we place cards so that the

center of gravity of the top k cards lies just above the edge of the

k-t

1st card

(which supports those top k). The table plays the role of the

1st card. To

express this condition algebraically, we can let

be the distance from the

extreme edge of the top card to the corresponding edge of the kth card from

the top. Then

= 0, and we want to make dk+, the center of gravity of the

first k cards:

&+l =

+l)+(dz+l)+...+(dk+l),

for1

<k<n

, .

(6,55)

260 SPECIAL NUMBERS

(The center of gravity of k objects, having respective weights

WI,

. . . ,

and having

reSpeCtiVe

Centers Of

gravity

pOSitiOnS

~1,

. . . pk, is at position

(WPl

’

WkPk)/bl

’

. +

wk).)

We can rewrite this recurrence in two

equivalent forms

k&+1

= k +

+ . . . +

dkp1

k 3 0;

(k-l)dk

= k-l

+dl

+...+dk-1, k>

Subtracting these equations tells us that

kdk+l

-(k-l)dk

+dk,

hence dk+l = dk + l/k. The second card will be offset half a unit past the

third, which is a third of a unit past the fourth, and so on. The general

formula

&+I

(6.56)

follows by induction, and if we set k = n we get dn+l =

as the total

overhang when n cards are stacked as described.

Could we achieve greater overhang by holding back, not pushing each

card to an extreme position but storing up “potential gravitational energy”

for a later advance? No; any well-balanced card placement has

&+I 6

(l+dl)+(l-td~)+...+(l+dk)

<k<n.

Furthermore dl = 0. It follows by induction that dk+l < Hk.

Notice that it doesn’t take too many cards for the top one to be com-

pletely past the edge of the table. We need an overhang of more than one

cardlength, which is 2 units. The first harmonic number to exceed 2 is

HJ =

so we need only four cards.

And with 52 cards we have an H52-unit overhang, which turns out to be

H52/2 x 2.27 cardlengths. (We will soon learn a formula that tells us how to

compute an approximate value of

for large n without adding up a whole

bunch of fractions.)

An amusing problem called the “worm on the rubber band” shows har-

monic numbers in another guise. A slow but persistent worm, W, starts at

one end of a meter-long rubber band and crawls one centimeter per minute

toward the other end. At the end of each minute, an equally persistent keeper

of the band, K, whose sole purpose in life is to frustrate W, stretches it one

meter. Thus after one minute of crawling, W is 1 centimeter from the start

and 99 from the finish; then K stretches it one meter. During the stretching

operation W maintains his relative position, 1% from the start and 99% from

Anyone who actu-

ally tries to achieve

this maximum

overhang with 52

cards is probably

not dealing with

a full deck-or

maybe he’s a real

joker.

6.3 HARMONIC NUMBERS 261

Metric units make

this problem more

scientific.

the finish; so W is now 2 cm from the starting point and 198 cm from the

goal. After W crawls for another minute the score is 3 cm traveled and 197

to go; but K stretches, and the distances become 4.5 and 295.5. And so on.

Does the worm ever reach the finish? He keeps moving, but the goal seems to

move away even faster. (We’re assuming an infinite longevity for K and W,

an infinite elasticity of the band, and an infinitely tiny worm.)

Let’s write down some formulas. When K stretches the rubber band, the

fraction of it that W has crawled stays the same. Thus he crawls l/lOOth of

it the first minute, 1/200th the second, 1/300th the third, and so on. After

n minutes the fraction of the band that he’s crawled is

100 (

1+!+1+

"'+n

)

100'

(6.57)

A flatworm, eh?

So he reaches the finish if

ever surpasses 100.

We’ll see how to estimate

for large

‘n

soon; for now, let’s simply

check our analysis by considering how “Superworm” would perform in the

same situation. Superworm, unlike W, can crawl 50cm per minute; so she

will crawl

HJ2

of the band length after n minutes, according to the argument

we just gave. If our reasoning is correct, Superworm should finish before n

reaches 4, since

> 2. And yes, a simple calculation shows that Superworm

has only 335 cm left to travel after three minutes have elapsed. She finishes

in 3 minutes and 40 seconds flat.

Harmonic numbers appear also in Stirling’s triangle. Let’s try to find a

closed form for

[‘J

, the number of permutations of n objects that have exactly

two cycles. Recurrence (6.8) tells us that

[“:‘I

[y]

=4 1

+(n-l)!,

ifn>O;

and this recurrence is a natural candidate for the summation factor technique

of Chapter 2:

n-t1

[

(n-l)!

+;.

Unfolding this recurrence tells us that

[nl’] =

H,;

hence

n+l

n!H,

(6.58)

We proved in Chapter 2 that the harmonic series

1 /k diverges, which

means that

gets arbitrarily large as n -+ 00. But our proof was indirect;

262 SPECIAL NUMBERS

we found that a certain infinite sum (2.58) gave different answers when it was

rearranged, hence ,Fk l/k could not be bounded. The fact that

seems counter-intuitive, because it implies among other things that a large

enough stack of cards will overhang a table by a mile or more, and that the

worm W will eventually reach the end of his rope. Let us therefore take a

closer look at the size of

when n is large.

The simplest way to see that

is probably to group its terms

according to powers of 2. We put one term into group 1, two terms into

group 2, four into group 3, eight into group 4, and so on:

1+1+

;+;+;:+;

~~‘~1~~~~~1~~~~

&-\I

8 9 10 11 12 13 14 15

group 1 group 2 group 3

group 4

Both terms in group 2 are between

and 5, so the sum of that group is

between 2.

and

= 1. All four terms in group 3 are between

and

so their sum is also between

and 1. In fact, each of the 2k-’ terms

in group k is between 22k and 21ek; hence the sum of each individual group

is between

and 1.

This grouping procedure tells us that if n is in group k, we must have

> k/2 and

6 k (by induction on k). Thus

co,

and in fact

LlgnJ

We now know

within a factor of 2. Although the harmonic numbers

approach infinity, they approach it only logarithmically-that is, quite slowly.

We should call them

Better bounds can be found with just a little more work and a dose

the

worm

numbers~

of calculus. We learned in Chapter 2 that

is the discrete analog of the

they’re so slow.

continuous function Inn. The natural logarithm is defined as the area under

a curve, so a geometric comparison is suggested:

f(x)

f(x) = l/x

0 1 2 3 . . . n

nfl

The area under the curve between 1 and n, which is Jy dx/x = Inn, is less

than the area of the n rectangles, which is

xF=:=,

l/k = H,. Thus Inn <

H,;

this is a sharper result than we had in (6.59). And by placing the rectangles

6.3 HARMONIC NUMBERS 263

‘7

now see a way

too

how

aggre-

gate of

termes

Musical1 pro-

gressions

may bee

found (much after

ye same manner)

Logarithms,

but

y”

calculations for

finding out those

rules would bee still

more troublesom.”

-1.

Newton [223]

a little differently, we get a similar upper bound:

0 1 2 3 . . . n

This time the area of the n rectangles, H,,

is less than the area of the first

rectangle plus the area under the curve. We have proved that

Inn <

< lnn+l, for n > 1.

(6.60)

We now know the value of

with an error of at most 1.

“Second order” harmonic numbers Hi2) arise when we sum the squares

of the reciprocals, instead of summing simply the reciprocals:

Hf’

,+;+;+...+$

x2.

k=l

Similarly, we define harmonic numbers of order r by summing (--r)th powers:

Ht)

f-&

k=l

(6.61)

If r >

these numbers approach a limit as n

--t

00; we noted in Chapter 4

that this limit is conventionally called Riemann’s zeta function:

(Jr) =

k>l

(6.62)

Euler discovered a neat way to use generalized harmonic numbers to

approximate the ordinary ones,

Let’s consider the infinite series

(6.63)

which converges when k > 1. The left-hand side is Ink

ln(k

1); therefore

if we sum both sides for 2 6 k 6 n the left-hand sum telescopes and we get

= (H,-1) +

;(HP’-1)

+ $(Hc’-1) +

;(H:)-1)

+ ... .

264 SPECIAL NUMBERS

Rearranging, we have an expression for the difference between

and Inn:

H,-Inn

= 1-

i(HF’

-1)

(j-$/%1)

$-$‘-1)

. . .

When n -+ 00, the right-hand side approaches the limiting value

-;&(2)-l)

-3&(3).-l)

$(LV-1)

-...

which is now known as Euler’s constant and conventionally denoted by the

Greek letter y. In fact, L(r)

1 is approximately

l/2’,

so this infinite series

“Huius

igitur

quan-

converges rather rapidly and we can compute the decimal value

titatis constantis

valorem

detex-

y = 0.5772156649. . . .

(6.64)

imus,

quippe est

C =

0,577218."

Euler’s argument establishes the limiting relation

lim (H,

-Inn)

n-CC

(6.65)

thus

lies about 58% of the way between the two extremes in (6.60). We

are gradually homing in on its value.

Further refinements are possible, as we will see in Chapter 9. We will

prove, for example, that

lnn+y+&--

1 2n2

+ 120n4

’

O<cn<l.

This formula allows us to conclude that the millionth harmonic

HIOOOOOO

14.3927267228657236313811275,

(6.66)

number is

without adding up a million fractions. Among other things, this implies that

a stack of a million cards can overhang the edge of a table by more than seven

cardlengths.

What does (6.66) tell us about the worm on the rubber band? Since

unbounded, the worm will definitely reach the end, when

first exceeds

100.

Our approximation to

says that this will happen when n is approximately

In fact, exercise 9.49 proves that the critical value of n is either [e’oo-‘J or

Well, they can ‘t

really go at it this

‘oo~~~l.

We can imagine W’s triumph when he crosses the finish line at last,

long;

the world will

much to K’s chagrin, some 287 decillion centuries after his long crawl began.

have ended much

(The rubber band will have stretched to more than

102’

light years long; its

earlier, when the

Tower of Brahma is

molecules will be pretty far apart.)

fully transferred.

6.4 HARMONIC SUMMATION 265

6.4 HARMONIC SUMMATION

Now let’s look at some sums involving harmonic numbers, starting

with a review of a few ideas we learned in Chapter 2. We proved in (2.36)

and (2.57) that

Hk =

O<k<n

kHk =

O<k<n

Let’s be bold and

nH,

-n;

(6.67)

n(n-

llH

n(n- 1)

lx-

(6.68)

take on a more general sum, which includes both of these

as special cases: What is the value of

when m is a nonnegative integer?

The approach that worked best for (6.67) and (6.68) in Chapter 2 was

called summation by parts. We wrote the summand in the form u(k)Av(k),

and we applied the general identity

~;u(x)Av(x)

u(x)v(x)(L

x:x(x

+ l)Au(x) 6x.

(6.69)

Remember? The sum that faces us now, xoSkcn (k)Hk, is a natural for this

method because we can let

u(k)

Hk,

Au(k) =

Hk+l

Hk =

;

Av(k) =

(In other words, harmonic numbers have a simple A and binomial coefficients

have a simple A-‘, so we’re in business.) Plugging into (6.69) yields

The remaining sum is easy, since we can absorb the (k + 1

)-’

using our old

standby, equation (5.5):

266 SPECIAL NUMBERS

Thus we have the answer we seek:

(&I,

OHk

(ml

(Hn-

$7).

(6.70)

(This checks nicely with (6.67) and (6.68) when m = 0 and m = 1.)

The next example sum uses division instead of multiplication: Let us try

to evaluate

f;.

k=l

If we expand Hk by its definition, we obtain a double sum,

Now another method from

us that

C:hapter

2 comes to our aid;

eqUatiOn

(2.33)

tdlS

k(($J2+g$)

~(H;+H?)).

(6.71)

It turns out that we could also have obtained this answer in another way if

we had tried to sum by parts (see exercise 26).

Now let’s try our hands at a more difficult problem

[291],

which doesn’t

submit to summation by parts:

integer n > 1

(This sum doesn’t explicitly mention harmonic numbers either; but who

(Not to give the

knows when they might turn up?)

answer away or

We will solve this problem in two ways, one by grinding out the answer

anything.)

and the other by being clever and/or lucky. First, the grinder’s approach. We

expand (n

k)” by the binomial theorem, so that the troublesome k in the

denominator will combine with the numerator:

;

(;)

(-k)jnn-j

k>l

(-l)i-lTln-j

(El)

(-l)kk’P’

k>l

This isn’t quite the mess it seems, because the

kj-’

in the inner sum is a

polynomial in k, and identity (5.40) tells us that we are simply taking the

6.4 HARMONIC SUMMATION

2ci7

nth difference of this polynomial. Almost; first we must clean up a few things.

For one, kim’ isn’t a polynomial if j = 0; so we will need to split off that term

and handle it separately. For another, we’re missing the term k = 0 from the

formula for nth difference; that term is nonzero when j =

so we had better

restore it (and subtract it out again). The result is

i>l

(-1)'

‘nnPix

(E)(-l)kki

’

k?O

OK, now the top line (the only remaining double sum) is zero: It’s the sum

of multiples of nth differences of polynomials of degree less than n, and such

nth differences are zero. The second line is zero except when j = 1, when it

equals

-nn.

So the third line is the only residual difficulty; we have reduced

the original problem to a much simpler sum:

(6.72)

For example,

Ll3

(:)$

(i)

= (:)

(:)

(:)i

= $$ hence

Ll3

27(T3

~ 1) as claimed.

How can we evaluate T,? One way is to replace

(F)

(“i’)

(:I:),

obtaining a simple recurrence for

T,,

in terms of T,

But there’s a more

instructive way: We had a similar formula in (5.41), namely

___

x(x+ l)...(x +

’

If we subtract out the term for k = 0 and set x = 0, we get

-Tn.

So let’s do it:

x(x+

1)::.

(x+n)

X=o

(x+l)...(x+n)-n!

(

x(x+l)...(x+n)

x=0

x”[~~~]

+...+x[“t’]

+ [n:‘] -n!=

(

x(x +

l)...

(x+

Ii0

;[“:‘I

268 SPECIAL NUMBERS

(We have used the expansion (6.11) of (x + 1) . . . (x + n) =

xn+‘/x;

we can

divide x out of the numerator because

[nt’]

= n!.) But we know from (6.58)

that

[nt’]

= n! H,; hence

T,,

= H,, and we have the answer:

Ll, =

n”(H,-1).

(6.73)

That’s one approach. The other approach will be to try to evaluate a

much more general sum,

U,(x,y) =

xG)‘g(~+ky)~,

integern30;

(6.74)

k>l

the value of the original Ll, will drop out as the special case U,(n, -1). (We

are encouraged to try for more generality because the previous derivation

“threw away” most of the details of the given problem; somehow those details

must be irrelevant, because the nth difference wiped them away.)

We could replay the previous derivation with small changes and discover

the value of U,(x,y). Or we could replace (x +

ky)”

by (x + ky)+‘(x + ky)

and then replace

(i)

by (“i’) +

(:I:),

leading to the recurrence

U,(x,y) =

xLLl(x,yj

+xn/n+yxn-’

;

(6.75)

this can readily be solved with a summation factor (exercise 5).

But it’s easiest to use another trick that worked to our advantage in

Chapter 2: differentiation. The derivative of

(x, y ) with respect to y brings

out a k that cancels with the k in the denominator, and the resulting sum is

trivial:

$.l,(x,

y) =

(1)

(-l)kP’n(x + ky)+’

k>l

nx”-’

(-l)kn(x

ky)nP’

nxnP’

(Once again, the nth difference of a polynomial of degree < n has vanished.)

We’ve proved that the derivative of U,(x, y) with respect to y is nxnP’,

independent of y. In general, if f’(y) = c then f(y) = f(0) + cy; therefore we

must have U,(x,y) = &(x,0) + nxnP’y.

The remaining task is to determine

(x,

0). But U,(x, 0) is just

times the sum

we’ve already considered in (6.72); therefore the

general sum in (6.74) has the closed form

Un(x, y) = xnHn +

nxnP’

y .

(6.76)

In particular, the solution to the original problem is

(n,

-1) = nn(Hn

1).

6.5 BERNOULLI NUMBERS 269

6.5

BERNOULLI NUMBERS

The next important sequence of numbers on our agenda is named

after Jakob Bernoulli

(1654-1705),

who discovered curious relationships while

working out the formulas for sums of mth powers

[22].

Let’s write

n-1

S,(n) =

Om+lm+...+(n-l)m

x;xmsx.

(6.77)

k=O

(Thus, when m > 0 we have S,(n) =

Hi::)

in the notation of generalized

harmonic numbers.) Bernoulli looked at the following sequence of formulas

and spotted a pattern:

So(n) = n

S,(n) = ?n

Sz(n)

= in3

in2 +

S3(n)

= in4

in3

+ in2

S4(n)

in5

in4 + in3

S5(n)

in6

+ fin4

+pz

!j6(n)

= +n’

in6

in5

in3

ST(n)

= in8

in’

+ An6

&n” + An2

&J(n)

in8

$n'-

&n5+

$n3-

ST(n)

&n’O

in9

$n8-

$n6+

$4-

&n2

So(n)

+lo+

in9-

n7+

n5-

1n3+5n

2 66

Can you see it too? The coefficient of

nm+’

in S,(n) is always 1 /(m + 1).

The coefficient of

is always

-l/2.

The coefficient of

nmP’

is always . . .

let’s see . . .

m/12. The coefficient of

nmP2

is always zero. The coefficient

of nmP3 is always . . . let’s see . . . hmmm . . . yes, it’s

-m(m-l)(m-2)/720.

The coefficient of

nmP4

is always zero. And it looks as if the pattern will

continue, with the coefficient of

nmPk

always being some constant times

mk.

That was Bernoulli’s discovery. In modern notation we write the coeffi-

cients in the form

S,(n)

&(Bcnmil

(m:l)B~nm+...+

(m~‘)Bmn)

(mk+‘)BkTlm+l-k.

k=O

(6.78)

270 SPECIAL NUMBERS

Bernoulli numbers are defined by an implicit recurrence relation,

B’

= [m==O], for all m 3 0.

For example, (i)Bo + (:)B’ = 0. The first few values turn out to be

(All conjectures about a simple closed form for

are wiped out by the

appearance of the strange fraction

-691/2730.)

We can prove Bernoulli’s formula (6.78) by induction on m, using the

perturbation method (one of the ways we found Sz(n) =

El,

in Chapter 2):

n-.1

,,,+I

(n) + nm+’ =

(k + l)m+’

k=O

(m:l)k’

(m:l)Sj(n).

(6.80)

Let

S,(n)

be the right-hand side of (6.78); we wish to show that S,,,(n) =

S,(n),

assuming that

(n) =

(n) for 0 < j < m. We begin as we did for

m = 2 in Chapter 2, subtracting

S,,,+’

(n) from both sides of (6.80). Then we

expand each

(n) using (6.78), and regroup so that the coefficients of powers

of n on the right-hand side are brought together and simplified:

nm+’

(m+l)Sj(,i

(mT1)5j(Tl) +

(“z’)

j=O

~(m~')~~~(jk')Bknj+l~'+~m+l)b

o~~~~(m~l)(i~l)~n’i’~~k+(m+l)A

o~~~,,(m~l)(~~~)~nk+l

+(m+l)A

6.5 BERNOULLI NUMBERS 271

Here’s some more

neat stuff that

you’ll probably

want to skim

through the first

time.

-Friend/y TA

Start

Skimming

o~,~(m~l)k~,(~~~k)Bj--r+(m+l)A

o~m~(m~l)o~~~i(m~~~k)~~+~~+~~A

. .

[m-k=Ol+(m+l)A

nm”

+ (m+

l)A,

where A = S,,,(n)

-g,(n).

(This derivation is a good review of the standard manipulations we learned

in Chapter 5.) Thus A = 0 and S,,,(n) =

S,(n),

QED.

In Chapter 7 we’ll use generating functions to obtain a much simpler

proof of (6.78). The key idea will be to show that the Bernoulli numbers are

the coefficients of the power series

(6.81)

Let’s simply assume for now that equation (6.81) holds, so that we can de-

rive some of its amazing consequences. If we add ;Z to both sides, thereby

cancelling the term

Blz/l!

-;z

from the right, we get

zeZ+l

eLi2

ecL12

-L+;

= -- =

eL-1

p/2

e-z/2

coth z

2 2’

(6.82)

Here coth is the “hyperbolic cotangent” function, otherwise known in calculus

books as

cash

z/sinh z; we have

sinhz =

e-2

ecz

coshz =

Changing

to --z gives

(7)

coth(

coth 5; hence every odd-numbered

coefficient of

coth

must be zero, and we have

= Bs = B, =

= B,, = B,3 =

...

= 0.

(6.84)

Furthermore (6.82) leads to a closed form for the coefficients of

coth:

zcothz =

-&+;

xB2,s

II>0

UP,,,&,

. (6.85)

nk0

But there isn’t much of a market for hyperbolic functions; people are more

interested in the “real” functions of trigonometry. We can express ordinary

272 SPECIAL NUMBERS

trigonometric functions in terms of their hyperbolic cousins by using the rules

sin z = -isinh

cos z =

cash

iz;

the corresponding power series are

sin2

2’

1!-3!+5!--...

sinhz =

T+“j-i.+5r+...;

cosz

o!-2!+4?--...)

.ci

coshz =

ol+2r+T+...

. . .

Hence cot

= cos z/sin

cash

iz/

sinh

= i

coth

iz, and we have

(6.86)

I see, we get “real”

functions

using

imaginary numbers.

(6.87)

Another remarkable formula for zcot z was found by Euler (exercise 73):

zcotz =

l-2tTg.

k>,krr

-z2

(6.88)

We can expand Euler’s formula in powers of

z2,

obtaining

Equating coefficients of

zZn

with those in our other formula,

(6.87),

gives us

an almost miraculous closed form for infinitely many infinite sums:

<(In)

H($)

(-l)np'

22n-1

n2nf3

(2n)!

’

integer n >

(6.89)

For example,

c(2)

HE)

;

n2B2

x2/6;

(6.90)

((4) = Hk) = 1 +

. . =

-ff

B4/3

d/90.

(6.91)

Formula (6.89) is not only a closed form for

HE),

it also tells us the approx-

imate size of

Bzn,

since H,,

(ln)

is very near 1 when n is large. And it tells

US that

(-l)n-l

B2,,

> 0 for all n > 0; thus the

nonzero

Bernoulli numbers

alternate in sign.

6.5 BERNOULLI NUMBERS 273

And that’s not all. Bernoulli numbers also appear in the coefficients of

the tangent function,

(6.92)

as well as other trigonometric functions (exercise 70). Formula

(6.92)

leads

to another important fact about the Bernoulli numbers, namely that

T2n-,

= (-1)-l

4n(4n-l)

Bzn

is a positive integer.

(Wi)

We have, for example:

135

7 9

Tll

1 2 16

272 7936 353792

22368256

(The T's are called tangent numbers.)

One way to prove

(6.g3),

following an idea of B. F. Logan, is to consider

the power series

sinz+xcosz

(l+x2)z+ (2x3+2x); + (6x4+8x2+2); +

cosz-xsinz

When x = tanw, where T,,(x) is a polynomial in x; setting x = 0 gives

(0) =

Tn,

the nth

this is tan( z + w) .

tangent number. If we differentiate

(6.94)

with respect to x, we get

(cosz-xsinz)2

xT(x)$;

Tl>O

but if we differentiate with respect to

we get

1+x2

(cosz-xsin~)~

tT,(xl&

tT,_M$.

ll>l tl)O

(Try it-the cancellation is very pretty.) Therefore we have

-&,+1(x)

+x2)T;(x),

To(x)

(fhd

a simple recurrence from which it follows that the coefficients of

Tn(x)

are

nonnegative integers. Moreover, we can easily prove that

Tn(x)

has degree

n + 1, and that its coefficients are alternately zero and positive. Therefore

Tz,+I

(0) =

Tin+,

is a positive integer, as claimed in

(6.93).

274 SPECIAL NUMBERS

Recurrence (6.95) gives us a simple way to calculate Bernoulli numbers,

via tangent numbers, using only simple operations on integers; by contrast,

the defining recurrence (6.79) involves difficult arithmetic with fractions.

If we want to compute the sum of nth powers from a to b

1 instead of

from 0 to n

1, the theory of Chapter 2 tells us that

b-l

k”’

xm6x = S,(b) -S,,,(a).

(6.96)

k=a

This identity has interesting consequences when we consider negative values

of k: We have

(-1)-F

km,

when m > 0,

k=--n+l

k=:O

hence

S,(O)

S,(-n+

(-l)m(Sm(n)

-S,(O)).

But S,(O) = 0, so we have the identity

S,(l

-n) =

(-l)“+‘S,(n),

m > 0.

(6.97)

Therefore

S,(

1) = 0. If we write the polynomial S,(n) in factored form, it

will always have the factors n and (n- 1 ), because it has the roots 0 and

general, S,(n) is a polynomial of degree m + 1 with leading term

&n”‘+’

Moreover, we can set n = i in (6.97) to get

S,(i)

(-l)“+‘S,(~);

if m is

even, this makes

S,(i)

= 0, so (n

5) will be an additional factor. These

observations explain why we found the simple factorization

Sl(n) =

in(n

t)(n

in Chapter 2; we could have used such reasoning to deduce the value of Sl(n)

without calculating it! Furthermore, (6.97) implies that the polynomial with

the remaining factors, S,(n) =

S,(n)/(n

i),

always satisfies

S,(l

-n) = S,(n), m even, m > 0.

It follows that S,(n) can always be written in the factored form

‘E’

;

ak)(n

;

Kk)

m odd;

S,(n)

k=l

(6.98)

6.5 BERNOULLI NUMBERS 275

Here

01’

and

0~2,

. . . ,

CX~,,,/~I

are appropriate complex numbers whose

values depend on m. For example,

Ss(n) =

n2(n-

1)2/4;

&t(n)

n(n-t)(n-l)(n-

t +

m)(n

fl)/5;

Ss(n) =

n’(n-l)‘(n-

m)(n-

m)/6;

Ss(n) =

n(n-$)(n-l)(n-i

(x)(n-5

Ix)(n--t

+E)(n-t

--I%),

where 01=

2~5i23~‘/231’i4(~~+

dm).

If m is odd and greater than 1, we have

= 0; hence S,,,(n) is divisible

(and by (n

1)‘). Otherwise the roots of S,(n) don’t seem to obey a

simple

law.

Let’s conclude our study of Bernoulli numbers by looking at how they

relate to Stirling numbers. One way to compute S,(n) is to change ordinary

powers to falling powers, since the falling powers have easy sums. After doing

those easy sums we can convert back to ordinary powers:

n-’

S,(n)

k=O

{;}l&

x{y}z

k=O j?O

j>O

k=O

t-11

j+l-

[

Therefore, equating coefficients with those in (6.78), we must have the identity

;{;}[i:‘](-jy;-*

--&(mk+l)Brn+i,.

(6.99)

It would be nice to prove this relation directly, thereby discovering Bernoulli

numbers in a new way. But the identities in Tables 250 or 251 don’t give

us any obvious handle on a proof by induction that the left-hand sum in

(6.99) is a constant times

rnc.

If k = m +

the left-hand sum is just

{R}

[EI;]/(m+l)

l/(m+l

so that case is easy. And if k = m, the

left-

handsidesumsto~~~,~[~]m~~-~~~[“‘~~](m+1~~~

=$(m-l)-im=-i;

so that case is pretty easy too. But if k < m, the left-hand sum looks hairy.

Bernoulli would probably not have discovered his numbers if he had taken

this route.

276 SPECIAL NUMBERS

Gnethingwecandoisreplace {y} by

{~~~}-(j+l){j~,}.

The

(j+l)

nicely cancels with the awkward denominator, and the left-hand side becomes

x.{~~"}[i;']~~

&{j;l}[i;'](-l)j+l-k

The second sum is zero, when k < m, by (6.31). That leaves us with the first

sum, which cries out for a change in notation; let’s rename all variables so

that the index of summation is k, and so that the other parameters are m

and n. Then identity (6.99) is equivalent to

{E}

[L]

“y”

~(~)B,-,,

[m=n-

11.

(6.100)

Good, we have something that looks more pleasant-although Table 251 still

doesn’t suggest any obvious next step.

The convolution formulas in Table 258 now come to the rescue. We can

use (6.51) and (6.50) to rewrite the summand in terms of Stirling polynomials:

(,-,)!hm(k);

%k(-k)

ok-m(k).

Things are looking good; the convolution in (6.48) yields

o,--k(-k) uk-,,,(k)

o,-,-k(-n

+ [n-m-k)) ok(m + k)

k=O k=O

(~l,l-“ni

unprn

(n-m))

Formula (6.100) is now verified, and we find that Bernoulli numbers are related

constant terms in the Stirling polynomials:

6.6

(-l)m~~‘mom(0) =

[m=l].

(6.101)

FIBONACCI NUMBERS

Now we come to a special sequence of numbers that is perhaps the

most pleasant of all, the Fibonacci sequence

(F,):

Fli

0 0 1 1 2 1 3 2 4 3 5 5

6 8

13 7 21 8

34 9

55 10 89 11

144 12 233 13 377 14

The back-to-nature

nature of this ex-

ample is shocking.

This book should be

banned.

Phyllotaxis, n.

The love of taxis.

6.6 FIBONACCI NUMBERS 277

Unlike the harmonic numbers and the Bernoulli numbers, the Fibonacci num-

bers are nice simple integers. They are defined by the recurrence

= 0;

= 1;

F,-I

+F,-2,

for n > 1.

(6.102)

The simplicity of this rule-the simplest possible recurrence in which each

number depends on the previous two-accounts for the fact that Fibonacci

numbers occur in a wide variety of situations.

“Bee trees” provide a good example of how Fibonacci numbers can arise

naturally. Let’s consider the pedigree of a male bee. Each male (also known

as a drone) is produced asexually from a female (also known as a queen); each

female, however, has two parents, a male and a female. Here are the first few

levels of the tree:

The drone has one grandfather and one grandmother; he has one great-

grandfather and two great-grandmothers; he has two great-great-grandfathers

and three great-great-grandmothers. In general, it is easy to see by induction

that he has exactly

Fn+l

greatn-grandpas and F,+z greatn-grandmas.

Fibonacci numbers are often found in nature, perhaps for reasons similar

to the bee-tree law. For example, a typical sunflower has a large head that

contains spirals of tightly packed florets, usually with 34 winding in one di-

rection and 55 in another. Smaller heads will have 21 and 34, or 13 and 21;

a gigantic sunflower with 89 and 144 spirals was once exhibited in England.

Similar patterns are found in some species of pine cones.

And here’s an example of a different nature

[219]:

Suppose we put two

panes of glass back-to-back. How many ways a,, are there for light rays to

pass through or be reflected after changing direction n times? The first few

278 SPECIAL NUMBERS

cases are:

a0 =

=2 az=3

When n is even, we have an even number of bounces and the ray passes

through; when n is odd, the ray is reflected and it re-emerges on the same

side it entered. The

a,‘s

seem to be Fibonacci numbers, and a little staring

at the figure tells us why: For n 3 2, the n-bounce rays either take their

first bounce off the opposite surface and continue in

a,-1

ways, or they begin

by bouncing off the middle surface and then bouncing back again to finish

a,-2

ways. Thus we have the Fibonacci recurrence a,, =

a,-1

a,-2.

The initial conditions are different, but not very different, because we have

= 1 =

and al = 2

F3;

therefore everything is simply shifted two

places, and a,, = F,+z.

Leonardo Fibonacci introduced these numbers in 1202, and mathemati-

cians gradually began to discover more and more interesting things about

them.

l%douard

Lucas, the perpetrator of the Tower of Hanoi puzzle dis-

cussed in Chapter 1, worked with them extensively in the last half of the

nine- “La suite de Fi-

teenth century (in fact it was Lucas who popularized the name “Fibonacci

bonacciPoss~de

numbers”). One of his amazing results was to use properties of Fibonacci

des propri&b

numbers to prove that the 39-digit Mersenne number

212’

1 is prime.

nombreuses

fort

inikkessantes.”

One of the oldest theorems about Fibonacci numbers, due to the French

astronomer Jean-Dominique Cassini in 1680

[45],

is the identity

-E. Lucas

[207]

,,+,F+,

-F; =

(-l).",

for n > 0.

(6.103)

When n = 6, for example, Cassini’s identity correctly claims that 1

3.5-tS2

A polynomial formula that involves Fibonacci numbers of the form

F,,+k

for small values of k can be transformed into a formula that involves only

and

F,+I

because we can use the rule

F,+2

F,+I

(6.104)

to express

in terms of higher Fibonacci numbers when m < n, and we can

use

F,~z+F,~,

(6.105)

to replace

by lower Fibonacci numbers when m > n-t1 . Thus, for example,

we can replace

F,-I

F,+I

in (6.103) to get Cassini’s identity in the

6.6 FIBONACCI NUMBERS 279

form

F:,,

F,+I

F,-F,f

= (-1)“. (6.106)

Moreover, Cassini’s identity reads

n+zFn

F,f+, =

(-l)“+’

when n is replaced by n + 1; this is the same as

(F,+I

F,)F,

F:,,

(-l)“+‘,

which is the same as (6.106). Thus Cassini(n) is true if and only if

Cassini(n+l) is true; equation (6.103) holds for all n by induction.

Cassini’s identity is the basis of a geometrical paradox that was one of

Lewis Carroll’s favorite puzzles

[54],

[258],

[298].

The idea is to take a chess-

board and cut it into four pieces as shown here, then to reassemble the pieces

into a rectangle:

Presto: The original area of 8 x 8 = 64 squares has been rearranged to yield

The paradox is

5 x 13 = 65 squares! A similar construction dissects any F, x F, square

explained

be-

cause well,

into four pieces, using

F,+I

F,,

and

as dimensions wherever the

magic tricks aren’t

illustration has 13, 8, 5, and 3 respectively. The result is an F, 1 x

F,+l

supposed to be

rectangle; by (6.103), one square has therefore been gained or lost, depending

explained.

on whether n is even or odd.

Strictly speaking, we can’t apply the reduction (6.105) unless m > 2,

because we haven’t defined F, for negative n. A lot of maneuvering becomes

easier if we eliminate this boundary condition and use (6.104) and (6.105) to

define Fibonacci numbers with negative indices. For example, F 1 turns out

to be

= 1; then F-

-F 1 = -1. In this way we deduce the values

0 -1 -2 -3 -4 -5 -6

-7

-8 -9 -10 -11

F, 1 0 1 -1 2 -3 5 -8 13 -21 34 -55 89

and it quickly becomes clear (by induction) that

Fm, =

(-l)nP’F,,

integer n.

(6.107)

Cassini’s identity (6.103) is true for all integers n, not just for n > 0, when

we extend the Fibonacci sequence in this way.

280 SPECIAL NUMBERS

The process of reducing

Fn*k

to a combination of F, and

F,+,

by using

(6.105) and (6.104) leads to the sequence of formulas

n+2

F,+I

Fn-I =

F,+I

n+3

2F,+,

Fn-2

-F,+,

+2F,

n+4

= 3F,+1 + 2F,

Fn-3

2F,+,

-SF,

n+5

= 5F,+1 + 3F,

Fn-4

-3F,+,

5F,

in which another pattern becomes obvious:

n+k

FkFn+l

h-IF,,

. (6.108)

This identity, easily proved by induction, holds for all integers k and n (pos-

itive, negative, or zero).

If we set k = n in

(6.108),

we find that

F2n =

FnFn+l

Fn-I

;

(6-g)

hence

Fz,,

is a multiple of F,. Similarly,

F3n = FznFn+tl

F2n-1Fn,

and we may conclude that

F:+,,

is also a multiple of F,. By induction,

Fkn

is a multiple of

(6.110)

for all integers k and n. This explains, for example, why

F15

(which equals

610) is a multiple of both

and

(which are equal to 2 and 5). Even more

is true, in fact; exercise 27 proves that

.wWm,

Fn)

Fgcd(m,n)

(6.111)

For example,

gcd(F,Z,F,s)

gcd(144,2584)

= 8 =

Fg.

We can now prove a converse of

(6.110):

If n

2 and if

is a multiple of

F,,

then m is a multiple of n. For if

F,\F,

then

F,\

gcd(F,, F,) =

Fgcd(m,n)

. .

F,. This

possible only if

Fgcd(m,nl

= F,; and our assumption that n > 2

makes it mandatory that gcd(m, n) = n. Hence n\m.

An extension of these divisibility ideas was used by Yuri Matijasevich in

his famous proof

[213]

that there is no algorithm to decide if a given multivari-

ate polynomial equation with integer coefficients has a solution in integers.

Matijasevich’s lemma states that, if n > 2, the Fibonacci number

is a

multiple of

if and only if m is a multiple of nF,.

Let’s prove this by looking at the sequence (Fk, mod

F$)

for k = 1, 2,

“‘,

and seeing when

Fk,,

mod

= 0. (We know that m must have the

6.6 FIBONACCI NUMBERS 281

form kn if

mod

= 0.) First we have

mod

F,;

that’s not zero.

Next we have

F2n

FnFn+l

F,-lF,

2F,F,+l

(mod

Fi)

by (6.108), since F,+I E

F,-l

(mod F,). Similarly

F2,+1 =

Fz+l +

E Fi+l (mod F,f).

This congruence allows us to compute

F3n

F2,+1

FznFn-I

= Fz+lF, +

(ZF,F,+I)F,+I

= 3Fz+,F,

(mod

Fi)

;

F3n+1

F2n+1

Fn+l

F2nFn

F;t+l

VFnF,+l

IF,

F:+l

(mod

F,f)

In general, we find by induction on k that

Fkn

E kF,F,k+; and Fk,,+l E

F,k+,

(mod

F:).

Now

Fn+l

is relatively prime to

F,,

Fkn

= 0 (mod Fz)

kF, E 0 (mod

F:)

k E 0 (mod F,).

We have proved Matijasevich’s lemma.

One of the most important properties of the Fibonacci numbers is the

special way in which they can be used to represent integers. Let’s write

j>>k

j 3

k+2.

(6.112)

Then every positive integer has a unique representation of the form

Fkz

Fk,

. . .

k, >> 0.

(6.113)

(This is “Zeckendorf’s theorem”

[201],

[312].) For example, the representation

of one million turns out to be

1~~0000

832040

121393

46368

144

F30

+ F26 +

F24

FIZ

+Flo.

We can always find such a representation by using a “greedy” approach,

choosing

Fk,

to be the largest Fibonacci number 6 n, then choosing

Fk2

to be the largest that is < n

Fk,, and so on. (More precisely, suppose that

282 SPECIAL NUMBERS

< n <

Fk+l;

then we have 0 6 n

Fk+l

Fk~

If n is a

Fibonacci number, (6.113) holds with

and kl = k. Otherwise n

has a Fibonacci representation

FkL

Fk,-,

by induction on n; and (6.113)

holds if we set kl = k, because the inequalities

FkL

< n

1 imply

that k

kz.) Conversely, any representation of the form (6.113) implies that

h,+l

because the largest possible value of

FkJ

+ . . . +

Fk,

when k

. . .

0 is

Fk~2$.Fk~4+...+FkmodZf2 =

Fk~m,

-1,

if k 3 2.

(6.114)

(This formula is easy to prove by induction on k; the left-hand side is zero

when k is 2 or 3.) Therefore

is the greedily chosen value described earlier,

and the representation must. be unique.

Any unique system of representation is a number system; therefore

Zeck-

endorf’s theorem leads to the Fibonacci number system. We can represent

any nonnegative integer n as a sequence of O’s and 1 ‘s, writing

n =

(b,b,-1

..bl)F

bkhc

(6.115)

k=2

This number system is something like binary (radix 2) notation, except that

there never are two adjacent

1's.

For example, here are the numbers from 1

to 20, expressed Fibonacci-wise:

(000001)~ 6

(OOIOO1)F

(010100)~

16 =

(lOOIOO)F

(000010)~

(001010)~

(010101)~

17=

(100101)~

(000100)~ 8

(OIOOOO)F

(100000)~

18 =

(lOIOOO)F

(000101)~

(010001)~ 14

(100001)~ 19

(101001)~

(001000)~ 10

(010010)~

(100010)~

20 =

(101010)~

The Fibonacci representation of a million, shown a minute

ago,

can be con-

trasted with its binary representation

219

218

2”

216

214

26:

(1000000)10

= (10001010000000000010100000000)~

(11110100001001000000)~.

The Fibonacci representation needs a few more bits because adjacent l's are

not permitted; but the two representations are analogous.

To add 1 in the Fibonacci number system, there are two cases: If the

“units digit” is 0, we change it to 1; that adds

= 1, since the units digit

6.6 FIBONACCI NUMBERS 283

5% 1

2xx

3x3

+5x4 +8x5

13x6

+21x'

34x8&c

Series

nata

divisione

Unitatis

per

Trinomium

1 -x-xx.”

-A. de Moivre [64]

“The quantities

r, s, t, which

show the relation

of the terms, are

the same as those in

the denominator of

the fraction. This

property, howsoever

obvious it may

be, M. DeMoivre

was the first that

applied it to use,

in the solution of

problems about

infinite series, which

otherwise would

have been very

intricate.”

-J.

Stirling

[281]

refers to Fz. Otherwise the two least significant digits will be 01, and we

change them to 10 (thereby adding

= 1). Finally, we must “carry”

as much as necessary by changing the digit pattern ‘011' to ‘100' until there

are no two l's in a row. (This carry rule is equivalent to replacing

Fm+l

+ F,

F,+z.)

For example, to go from 5 = (1000)~ to 6 =

(1001)~

or from

6 = (1001 )r to 7 =

(1010)~

requires no carrying; but to go from 7 =

(1010)~

to 8 =

(1OOOO)r

we must carry twice.

So far we’ve been discussing lots of properties of the Fibonacci numbers,

but we haven’t come up with a closed formula for them. We haven’t found

closed forms for Stirling numbers, Eulerian numbers, or Bernoulli numbers

either; but we were able to discover the closed form

[“:‘]/n!

for har-

monic numbers. Is there a relation between

and other quantities we know?

Can we “solve” the recurrence that defines F,?

The answer is yes. In fact, there’s a simple way to solve the recurrence by

using the idea of generating

finction

that we looked at briefly in Chapter 5.

Let’s consider the infinite series

F(z)

F1:z+

Fzz2

+...

= tF,,z". (6.116)

TX20

If we can find a simple formula for F(z), chances are reasonably good that we

can find a simple formula for its coefficients

F,.

In Chapter 7 we will focus on generating functions in detail, but it will

be helpful to have this example under our belts by the time we get there.

The power series

F(z)

has a nice property if we look at what happens when

we multiply it by z and by z2:

F(z) =

Flz

F2z2

F3z3

Fqz4

F5z5

...

zF(z)

Fez

F,z2

F2z3

F3z4

F4z5

...

z'F(z)

Foz2

F,z3

F2z4

F3z5

...

If we now subtract the last two equations from the first, the terms that involve

z2,

23,

and higher powers of z will all disappear, because of the Fibonacci

recurrence. Furthermore the constant term

never actually appeared in the

first place, because

= 0. Therefore all that’s left after the subtraction is

(F,

Fg)z,

which is just z. In other words,

F(z)-zF(z)-z.zF(z)

and solving for F(z) gives us the compact formula

F(z) =

L-.

l-Z-22

(6.117)

284 SPECIAL NUMBERS

We have now boiled down all the information in the Fibonacci sequence

to a simple (although unrecognizable) expression z/( 1

2’).

This, believe

it or not, is progress, because we can factor the denominator and then use

partial fractions to achieve a formula that we can easily expand in power series.

The coefficients in this power series will be a closed form for the Fibonacci

numbers.

The plan of attack just sketched can perhaps be understood better if

we approach it backwards. If we have a simpler generating function, say

l/(

az)

where

is a constant, we know the coefficients of all powers of z,

because

-az

1+az+a2z2+a3z3+~~~.

Similarly, if we have a generating function of the form A/( 1

az)

B/(

pz),

the coefficients are easily determined, because

+1+3z

A~(az)"+B~(@)"

1120 ll?O

Aa” + BBn)z” .

(6.118)

n>o

Therefore all we have to do is find constants A, B, a, and

such that

B z

t-m=

~~~

and we will have found a closed form Aa” + BP” for the coefficient F, of

z”

in F(z). The left-hand side can be rewritten

A-A@+B-Baz

-az

+1-f3z

Il-az)(l-pz)

’

so the four constants we seek are the solutions to two polynomial equations:

-az)(l

-f32) = 1

-z-z2;

(6.119)

(A-t-B)-(A@+Ba)z

= z.

(6.120)

We want to factor the denominator of

F(z)

into the form (1

az)(l

(3~);

then we will be able to express

F(z)

as the sum of two fractions in which the

factors (1

az) and (1

Bz)

are conveniently separated from each other.

Notice that the denominator factors in

(6.119)

have been written in the

form

az)

(3z),

instead of the more usual form c(z

~1)

~2)

where

and

are the roots. The reason is that (1

az)( 1

/3z)

leads to nicer

expansions in power series.

6.6 FIBONACCI NUMBERS 285

As usual, the au-

thors can't resist

a trick.

The

ratio

of one’s

height to the height

of one’s nave/ is

approximate/y

1.618, accord-

ing

to extensive

empirical observa-

tions by European

scholars

[ll

O].

We can find

,and

in several ways, one of which uses a slick trick: Let

us introduce a new variable w and try to find the factorization

w=-wz-z2

cxz)(w

bz) .

Then we can simply set w = 1 and we’ll have the factors of 1

z2.

The

roots of w2

= 0 can be found by the quadratic formula; they are

z*dJz2+4zz

1+Js

2=.

Therefore

-wz-z=

(

l+dS

1-d

w--z

)

and we have the constants

and

we were looking for.

The number (1 +

fi)/2

= 1.61803 is important in many parts of mathe-

matics as well as in the art world, where it has been considered since ancient

times to be the most pleasing ratio for many kinds of design. Therefore it

has a special name, the golden ratio. We denote it by the Greek letter

c$,

honor of Phidias who is said to have used it consciously in his sculpture. The

other root (1

fi)/2

-l/@

.61803 shares many properties of

so it

has the special name

“phi hat!’ These numbers are roots of the equation

w2-w-l

=O,sowehave

c$2

@+l;

$+l.

(More about

cj~

and $ later.)

(6.121)

We have found the constants

and

needed in (6.119); now

we merely need to find A and B in (6.120). Setting

= 0 in that equation

tells us that B = -A, so (6.120) boils down to

-$A+@A = 1.

The solution is A = 1

/(c$

= 1

/fi;

the partial fraction expansion of

(6.117) is therefore

Good, we’ve got F(z) right where we want it. Expanding the fractions into

power series as in (6.118) gives a closed form for the coefficient of

zn:

4”).

('5.123)

(This formula was first published by Leonhard Euler

[91]

in 1765, but people

forgot about it until it was rediscovered by Jacques Binet

[25]

in 1843.)

286 SPECIAL NUMBERS

Before we stop to marvel at our derivation, we should check its accuracy.

For n = 0 the formula correctly gives

= 0; for n = 1, it gives

9)/v%, which is indeed 1. For higher powers, equations (6.121) show

that the numbers defined by (6.123) satisfy the Fibonacci recurrence, so they

must be the Fibonacci numbers by induction. (We could also expand 4”

and $” by the binomial theorem and chase down the various powers of

but that gets pretty messy. The point of a closed form is not necessarily to

provide us with a fast method of calculation, but rather to tell us how

relates to other quantities in mathematics.)

With a little clairvoyance we could simply have guessed formula (6.123)

and proved it by induction. But the method of generating functions is a pow-

erful way to discover it; in Chapter 7 we’ll see that the same method leads us

to the solution of recurrences that are considerably more difficult. Inciden-

tally, we never worried about whether the infinite sums in our derivation of

(6.123) were convergent; it turns out that most operations on the coefficients

of power series can be justified rigorously whether or not the sums actually

converge

[151].

Still, skeptical readers who suspect fallacious reasoning with

infinite sums can take comfort in the fact that equation

(6.123),

once found

by using infinite series, can be verified by a solid induction proof.

One of the interesting consequences of (6.123) is that the integer F, is

extremely close to the irrational number I$~/& when n is large. (Since $ is

less than 1 in absolute value, $” becomes exponentially small and its effect

is almost negligible.) For example,

Flo

= 55 and

F11

= 89 are very near

55.00364

and

88.99775.

We can use this observation to derive another closed form,

rounded to the nearest integer,

(6.124)

because ( Gn/& 1<

for all. n 3 0. When n is even, F, is a little bit less

than

+“/&;

otherwise it is

little greater.

Cassini’s identity (6.103) can be rewritten

n+l

Fll

(-1

)T'

---=--

Fn-I Fn-I

When n is large, 1 /F,-1 F, is very small, so

F,,+l

/F, must be very nearly the

same as

F,/F,-I;

and (6.124) tells us that this ratio approaches 4. In fact,

we have

n+l

$F,

$”

(6.125)

If the USA ever

goes metric, our

speed limit signs

will go from 55

mi/hr to 89 km/hr.

Or maybe the high.

way people

will

generous and let us

go 90.

The “shift down”

rule changes n

f(n/@)

and

the “shift

up”

rule changes n

to f

(n+)

, where

f(x) =

+ @‘J

6.6 FIBONACCI NUMBERS 287

(This identity is true by inspection when n = 0 or n =

and by induction

when n > 1; we can also prove it directly by plugging in (6.123).) The ratio

F,+,/F,

is very close to

which it alternately overshoots and undershoots.

By coincidence,

is also very nearly the number of kilometers in a mile.

(The exact number is 1.609344, since 1 inch is exactly 2.54 centimeters.)

This gives us a handy way to convert mentally between kilometers and miles,

because a distance of

F,+l

kilometers is (very nearly) a distance of

miles.

Suppose we want to convert a non-Fibonacci number from kilometers

to miles; what is 30 km, American style? Easy: We just use the Fibonacci

number system and mentally convert 30 to its Fibonacci representation 21 +

8 + 1 by the greedy approach explained earlier. Now we can shift each number

down one notch, getting 13 + 5 + 1. (The former '1' was

Fz,

since k,

0 in

(6.113); the new ‘1’ is Fl.) Shifting down divides by

more or less. Hence

19 miles is our estimate. (That’s pretty close; the correct answer is about

18.64 miles.) Similarly, to go from miles to kilometers we can shift up a

notch; 30 miles is approximately 34 + 13 + 2 = 49 kilometers. (That’s not

quite as close; the correct number is about 48.28.)

It turns out that this “shift down” rule gives the correctly rounded num-

ber of miles per n kilometers for all n < 100, except in the cases n = 4, 12,

62, 75, 91, and 96, when it is off by less than 2/3 mile. And the “shift up”

rule gives either the correctly rounded number of kilometers for n miles, or

1 km too

mariy,

for all n < 126. (The only really embarrassing case is n = 4,

where the individual rounding errors for n = 3 + 1 both go the same direction

instead of cancelling each other out.)

6.7 CONTINUANTS

Fibonacci numbers have important connections to the Stern-Brocot

tree that we studied in Chapter 4, and they have important generalizations to

a sequence of polynomials that Euler studied extensively. These polynomials

are called continuants, because they are the key to the study of continued

fractions like

(6.126)

a2 +

a3 +

a5 +

___

a6 +

288 SPECIAL NUMBERS

The continuant polynomial

K,(x1

,x2,.

. . , x,) has n parameters, and it

is defined by the following recurrence:

KoO

;

(xl) =

;

&(x1,.

,x,)

= Kn-1

(xl,.

. .

,x,-l

)x,

Kn-2(x1,. . . ,

~-2).

(6.127)

For example, the next three cases after

(x1)

are

Kz(x1 ,x2)

x1x2

;

K3(xl,x2,x3)

x1x2x3+x1 +x3;

K4(xl,x2,x3,x4)

xlx2x3x4+x1x2+xlx4+x3x4+~

It’s easy to see, inductively, that the number of terms is a Fibonacci number:

K,(l,l,...

,I) =

Fn+l

(6.128)

When the number of parameters is implied by the context, we can write

simply ‘K’ instead of ‘K,‘,

,just

as we can omit the number of parameters

when we use the hypergeometric functions F of Chapter 5. For example,

K(x1,

x2)

Kz(xl

x2)

+ 1. The subscript n is of course necessary in

formulas like

(6.128).

Euler observed that

K(x1,

x2,

. . .

,x,,)

can be obtained by starting with

the product

. . .

x,,

and then striking out adjacent pairs

xkXk+l

in all

possible ways. We can represent Euler’s rule graphically by constructing all

“Morse code” sequences of dots and dashes having length n, where each dot

contributes 1 to the length and each dash contributes 2; here are the Morse

code sequences of length 4:

..-

.-.

-..

These dot-dash patterns correspond to the terms of

K(xl

,x2,x3, x4); a dot

signifies a variable that’s included and a dash signifies a pair of variables

that’s excluded. For example, l

l corresponds to

x1x4.

A Morse code sequence of length n that has k dashes has n-2k dots and

k symbols altogether. These dots and dashes can be arranged in (“i”)

ways; therefore if we replace each dot by z and each dash by 1 we get

K,,(z,

z,. .

PZk

(6.129)

6.7 CONTINUANTS 289

We also know that the total number of terms in a continuant is a Fibonacci

number; hence we have the identity

F,,+I

(“;

“)

k=O

(6.130)

(A closed form for (6.12g), generalizing the Euler-Binet formula (6.123) for

Fibonacci numbers, appears in (5.74).)

The relation between continuant polynomials and Morse code sequences

shows that continuants have a mirror symmetry:

K(x,,

x2,x1) =

K(x1,xr,...,xn).

(6.131)

Therefore they obey a recurrence that adjusts parameters at the left, in ad-

dition to the right-adjusting recurrence in definition (6.127):

K,(xI,...

,%I) =

XI&

1(X2,...,&1)

+Kn

2(x3,...,&).

(6.132)

Both of these recurrences are special cases of a more general law:

m+*(X1,...,X,,X,+1,~..,x~+~)

K,(xl,...,x,)K,(x,+~,...,x,+,)

+kn

I(xI,...,x,

l)K,

1(~,+2,...,~rn+n).

(6.133)

This law is easily understood from the Morse code analogy: The first product

K,K,

yields the terms of

K,+,

in which there is no dash in the [m, m +

position, while the second product yields the terms in which there is a dash

there. If we set all the x’s equal to 1, this identity tells us that Fm+n+l =

Fm+lF,+l +

F,F,;

thus, (6.108) is a special case of (6.133).

Euler

[90]

discovered that continuants obey an even more remarkable law,

which generalizes Cassini’s identity:

m+n(Xlr.~~

Xm+n)

Kk(Xm+l,

. . . ,

%n+k)

kn+k(Xl,

. . .

rX,+k)K,(x,+l,...,x,+,)

(-l)kKm

I(XI,...,X,

l)Kn

1(%n+k+2,...,Xm+,).

(6.134)

This law (proved in exercise 29) holds whenever the subscripts on the K’s are

all nonnegative. For example, when k = 2, m =

and n = 3, we have

K(xl,x2,x3,x4)K(x2,x3)

= K(Xl,X2,X?,)K(XL,X3,X4)

Continuant polynomials are intimately connected with Euclid’s algo-

rithm. Suppose, for example, that the computation of gcd(m, n) finishes

290 SPECIAL NUMBERS

in four steps:

@Cm,

gcd(no,

nl 1

= gcd(nl ,

= gcd(nr,n3’l

= gcd(n3,

na‘i

= gcd(ns,O) =

= m, nl

=n;

nomodn,

no-qlnl;

= nl mod

= nl

q2n2

;

nzmodn3

nz-q3n3;

0 =

modn4 =

q4n4.

Then we have

= K()n4

;

q4n4

K(q4h;

+n4 =

K(q3,q4h;

q2n2

+n3 = K(qZlq3,q4)n4;

qlnl

K(ql,q2,q3,q4h

In general, if Euclid’s algorithm finds the greatest common divisor d in k steps,

after computing the sequence of quotients ql, . . . , qk, then the starting num-

bers were K(ql,qz,.. . ,qk)d and K(q2,. . . ,

qk)d. (This fact was noticed early

in the eighteenth century by Thomas Fantet de Lagny

[190],

who seems to

have been the first person to consider continuants explicitly. Lagny pointed

out that consecutive Fibonacci numbers, which occur as continuants when the

q’s take their minimum values, are therefore the smallest inputs that cause

Euclid’s algorithm to take a given number of steps.)

Continuants are also intimately connected with continued fractions, from

which they get their name. We have, for example,

a0 +

K(ao,al,az,a3)

-K(al,az,a3)

(6.135)

a2 +

The same pattern holds for continued fractions of any depth. It is easily

proved by induction; we have, for example,

K(ao,al,az,a3+l/a4)

K(ao,al,a2,a3,a4)

K(al, az,

+ l/a41

K(al,az,as,ad)

’

because of the identity

K,(xl,.

. .

,xn-lrxn+Y)

K,(x,,...

,xn~l,x,)+Kn-l(xl,...,xn~l)~

(This identity is proved and generalized in exercise 30.)

(6.136)

6.7 CONTINUANTS 291

Moreover, continuants are closely connected with the Stern-Brocot tree

discussed in Chapter 4. Each node in that tree can be represented as a

sequence of L’s and

R'S,

say

RQO

La’

R”Z

L”’

. . .

Ran-’

LO-“-’

(6.137)

where

0, al

1, a3

. . . ,

a,-2

an 1 3

and n

even. Using the 2 x 2 matrices L and

of (4.33), it is not hard to prove by

induction that the matrix equivalent of (6.137) is

K,-2(al,.

)

an-21

Kn-l(al,...,an-2,an

K,-l(ao,al,...,an-2)

Kn(ao,al,...,an~~2,an~l)

(The proof is part of exercise 80.) For example,

R”LbRcLd

bc + 1 bcd+b+d

abc + a + c

abcd+ab+ad+cd+l

Finally, therefore, we can use (4.34) to write a closed form for the fraction in

the Stern-Brocot tree whose L-and-R representation is (6.137):

f(R""

.L"-')

Kn+l(ao,al,...~an~l,l)

K,(al,.

an-l,

(6.139)

(This is “Halphen’s theorem” [143].) For example, to find the fraction for

LRRL we have

= 0,

= 2,

= 1, and n = 4; equation (6.13~)

gives

K(O,

1,&l,

KC4

1,l)

U&2)

K(l,Ll,l)

K(1,2,1,1)

=-=-

K(3,2) 7

’

(We have used the rule

K,(xl,.

. .

,x,-l,

K,+, (XI,. . .

,x,-r

,x,,,

absorb leading and trailing l’s in the parameter lists; this rule is obtained by

setting y = 1 in (6.136).)

A comparison of (6.135) and (6.13~) shows that the fraction correspond-

ing to a general node (6.137) in the Stern-Brocot tree has the continued

fraction representation

f(Rao..

Lo-+’

) =

(6.140)

al +

I+-

292 SPECIAL NUMBERS

Thus we can convert at sight between continued fractions and the correspond-

ing nodes in the Stern-Brocot tree. For example,

f(LRRL) = 0+

l+-7

2 $-

1,;

We observed in Chapter 4 that irrational numbers define infinite paths

in the Stern-Brocot tree, and that they can be represented as an infinite

string of L’s and R’s. If the infinite string for a is

RaoLal

RaZL”3

. . . , there is

a corresponding infinite continued fraction

a = aof

(‘3.141)

a2 +

a4 +

a5 +

This infinite continued fraction can also be obtained directly: Let

= a and

for k 3 0 let

Lakj

;

ak+-.

Kkfl

(6.142)

The a’s are called the “partial quotients” of a. If a is rational, say m/n,

this process runs through the quotients found by Euclid’s algorithm and then

stops (with akfl =

o0).

Is Euler’s constant y rational or irrational? Nobody knows. We can get

Or if

they do,

partial information about this famous unsolved problem by looking for y in

theY’re

not

ta’king.

the Stern-Brocot tree; if it’s rational we will find it, and if it’s irrational we

will find all the closest rational approximations to it. The continued fraction

for y begins with the following partial quotients:

Therefore its Stern-Brocot representation begins LRLLRLLRLLLLRRRL . .

;

pattern is evident. Calculations by Richard Brent

[33]

have shown that, if y

is rational, its denominator must be more than 10,000 decimal digits long.

6.7 CONTINUANTS 293

Well, y must be

irrational, because

of a little-known

Einsteinian asser-

tion: “God does

not throw huge

denominators at

the universe.”

Therefore nobody believes that y is rational; but nobody so far has been able

to prove that it isn’t.

Let’s conclude this chapter by proving a remarkable identity that ties a lot

of these ideas together. We introduced the notion of spectrum in Chapter 3;

the spectrum of

is the multiset of numbers

Ln&],

where

is a given constant.

The infinite series

can therefore be said to be the generating function for the spectrum of

where

= (1 +

fi)/2

is the golden ratio. The identity we will prove, dis-

covered in 1976 by J.L. Davison

[61],

is an infinite continued fraction that

relates this generating function to the Fibonacci sequence:

(6.143)

Both sides of (6.143) are interesting; let’s look first at the numbers

Ln@J.

If the Fibonacci representation (6.113) of n is

Fk,

+ . . . + Fk,, we expect n+

to be approximately

Fk,

. . + Fk,+i , the number we get from shifting the

Fibonacci representation left (as when converting from miles to kilometers).

In fact, we know from (6.125) that

Fk,+,

Fk,+l

($“I

q”r)

Now+=-l/@andki

>...>>k,>>O,sowehave

and qkl

+..

.+$jkl has the same sign as (-1) kr, by a similar argument. Hence

In+] = Fk,+i

+.‘.+Fk,+l

[k,(n) iseven]. (6.144)

Let us say that a number n is Fibonacci odd (or F-odd for short) if its least

significant Fibonacci bit is 1; this is the same as saying that k,(n) = 2.

Otherwise n is Fibonacci even (F-even). For example, the smallest F-odd

294 SPECIAL NUMBERS

numbers are

6, 9, 12,

14, 17,

and

19.

If k,(n)

is even,

then

F-even, by (6.114); similarly, if k,(n) is odd, then n

1 is F-odd. Therefore

k,(n) is even

1 is F-even.

Furthermore, if k,(n) is even, (6.144) implies that

kT(

[n+])

= 2; if k,(n) is

odd, (6.144) says that

kr(

[rt@]) = k,(n) + 1. Therefore k,.( [n+J) is always

even, and we have proved that

In@]

1 is always F-even.

Conversely, if m is any F-even number, we can reverse this computation and

find an n such that m + 1

Ln@J.

(First add 1 in F-notation as explained

earlier. If no carries occur, n is (m + 2) shifted right; otherwise n is (m + 1)

shifted right.) The right-hand sum of (6.143) can therefore be written

LQJ

zm [m is F-even] ,

TL>l

ll@O

(6.145)

How about the fraction on the left? Let’s rewrite (6.143) so that the

continued fraction looks like (6.141), with all numerators 1:

1-Z

-=-

zcFfi

,lMJ

z-h

lI>l

z-F2 +

(6.146)

(This transformation is a bit tricky! The numerator and denominator of the

original fraction having

zFn

as numerator should be divided by

zFnmI

we stop this new continued fraction at l/zPFn, its value will be a ratio of

continuants,

K,,.z(O,

2~~0,

zPFI,. . .

,zPFn)

K,(z/

, . . . ,

z-~,)

K,+, (z-~o,z~~I,. . .

,zpFn)

K,+, (z-~o,

z-~I,.

. ,

z-~,)

’

as in (6.135). Let’s look at the denominator first, in hopes that it will be

tractable. Setting Qn = K,+l z

Fo,.

. ,zPFn), we find Q. = 1, Q, = 1 +

z-l,

-tz--’ + -2 Q

‘-I

z + z-2 +

zP3

+ zP4, and in general everything

fits beautifully and gives a geometric series

Q,, =

z-’

+ z-2 + . . . + z-(Fn+2-l 1 .

6.7 CONTINUANTS 295

The corresponding numerator is P, =

K,(zpF’,

. . . ,

zpFn);

this turns out to

be like

Q,,

but with fewer terms. For example, we have

compared with

= 1 +

z-'

z--12.

A closer look reveals the pattern

governing which terms are present: We have

= 1 +22+z3+z5+z7+z8+z’o+z”

Z’2

z-12

is F-even]

;

m=O

and in general we can prove by induction that

F,+z-’

z’-Fn+~

is F-even]

m=O

Therefore

Pll

t’,“Ji-’

z”’ [m is F-even]

QTI

xLL;p’ Zm

’

Taking the limit as n

0;)

now gives

(6.146),

because of (6.145).

Exercises

Warmups

What are the

[i]

= 11 permutations of

,2,3,4} that have exactly two

cycles? (The cyclic forms appear in (6.4); non-cyclic forms like 2314 are

desired instead.)

There are mn functions from a set of n elements into a set of m elements.

How many of them range over exactly k different function values?

Card stackers in the real world know that it’s wise to allow a bit of slack

so that the cards will not topple over when a breath of wind comes along.

Suppose the center of gravity of the top k cards is required to be at least

units from the edge of the k + 1st card. (Thus, for example, the first

card can overhang the second by at most

-c

units.) Can we still achieve

arbitrarily large overhang, if we have enough cards?

Express

l/l +

l/3

+...

1/(2n+l)

in terms of harmonic numbers.

Explain how to get the recurrence (6.75) from the definition of

L&,(x,

(6.74),

and solve the recurrence.

296 SPECIAL NUMBERS

6 An explorer has left a pair of baby rabbits on an island. If baby rabbits

become adults after one month, and if each pair of adult rabbits produces

one pair of baby rabbits every month, how many pairs of rabbits are

present after n months’? (After two months there are two pairs, one of

which is newborn.) Find a connection between this problem and the “bee

tree” in the text.

7 Show that Cassini’s identity (6.103) is a special case of (6.108), and a

special case of (6.134).

8 Use the Fibonacci number system to convert 65 mi/hr into an approxi-

mate number of km/hr.

9 About how many square kilometers are in 8 square miles?

What is the continued fraction representation of

Basics

11 What is

I:,(-l)“[t],

row sum of Stirling’s cycle-number triangle

with alternating signs, when n is a nonnegative integer?

12 Prove that Stirling numbers have an inversion law analogous to (5.48):

g(n) =

{t}(--1

lkf(k)

f(n) =

[L]

(-l)kg(k).

13 The differential operators D = & and 4 = zD are mentioned in Chapters

2 and 5. We have

= z2D2+zD,

because

a2f(z)

&f’(z)

= z&zf’(z) =

z2f”(z)

+ zf’(z), which is

(z2D2+zD)f(z). Similarly it can be shown that

= z3D3+3z2D2+zD.

Prove the general formulas

for all n 3 0. (These can be used to convert between differential expres-

sions of the forms

cxkzkfik’(z)

and

fikakf(z), as in (5.1og).)

14 Prove the power identity (6.37) for Eulerian numbers.

15 Prove the Eulerian identity (6.39) by taking the mth difference of (6.37).

6 EXERCISES 297

16 What is the general solution of the double recurrence

n,O = %

[n>ol

;

Ao,k

= 0,

ifk>O;

n.k

k&-l,k

A,-

l,k-1

integers k, n,

when k and n range over the set of all integers?

17 Solve the following recurrences, assuming that

I;/

is zero when n < 0 or

k < 0:

IL1

/n~l~+nl~~~l+[~~=k=Ol,

for n, k > 0.

b /;I =

(n--

k)lnkl/

lLz:l

[n=k=Ol,

for n, k 3 0.

c I;/ =

k~n~l~+k~~~~~+[n=k=O],

for n, k 3 0.

18 Prove that the Stirling polynomials satisfy

(x+l)~n(x+l)

(x-n)o,(x)+xo,-,(x)

19 Prove that the generalized Stirling numbers satisfy

~{x~k}[xe~+k](-l)k/(~‘+:)

intewn>O.

[x~k]{x~~+k}i-lik/(~++:)

integern>O.

20 Find a closed form for

xz=,

Hf’.

21 Show that if

= an/bn, where a, and b, are integers, the denominator

b, is a multiple of

2L1snj.

Hint: Consider the number

2L1snl

-‘H,

22 Prove that the infinite sum

converges for all complex numbers z, except when

is a negative integer;

and show that it equals H, when

is a nonnegative integer. (Therefore we

can use this formula to define harmonic numbers H, when

is complex.)

23 Equation (6.81) gives the coefficients of z/(e’

1), when expanded in

powers of z. What are the coefficients of z/(e’ + 1 )? Hint: Consider the

identity

(e’+

l)(e’-

1) =

ezZ-

298 SPECIAL NUMBERS

24 Prove that the tangent number Tz,+l is a multiple of 2”. Hint: Prove

that all coefficients of Tz,,(x) and

Tzn+l

(x) are multiples of 2”.

25 Equation (6.57) proves that the worm will eventually reach the end of

the rubber band at some time N. Therefore there must come a first

time n when he’s closer to the end after n minutes than he was after

1 minutes. Show that n <

:N.

26 Use summation by parts to evaluate

xr=,

Hk/k. Hint: Consider

also the related sum

Et=,

Hk-r/k.

2’7 Prove the gcd law

(6.111)

for Fibonacci numbers.

28 The Lucas number

is defined to be

Fn+r

+ F,--r. Thus, according to

(6.log), we have

Fzn

F,L,.

Here is a table of the first few values:

0 1 2 3 4 5 6 7 8 9 10 11 12 13

L,,I 2

29 47 76

123

199 322

521

Use the repertoire method to show that the solution Qn to the gen-

eral recurrence

Qo = a;

= B;

Qn-l+Qn-2,

n>l

can be expressed in terms of

and L,.

b Find a closed form for

in terms of 4 and

29 Prove Euler’s identity for continuants, equation (6.134).

30 Generalize (6.136) to find an expression for the incremented continuant

K(x,, . . .

,~,,~l,~~+y,~~+l,...,

x,,),

when

m<n.

Homework exercises

31 Find a closed form for the coefficients [:I in the representation of rising

powers by falling powers:

y=xl

kk'

integer n > 0.

(For example,

x4=x%+

12x3+36x2+24x1,

hence

141

36.).

32 In Chapter 5 we obtained the formulas

&(“:“)

(n+mm+l)

and

o&m(:)

(:I:)

by unfolding the recurrence

(c)

(“i’)

(:I:)

in two ways. What

identities appear when the analogous recurrence {L} = k{

“i’

}

{

:I,’

}

is unwound?

6 EXERCISES

299

33 Table 250 gives the values of

[;I

and

{

;} What are closed forms (not

involving Stirling numbers) for the next cases,

[;]

and

{‘;}?

34 What are

(:)

and

(-,‘),

if the basic recursion relation (6.35) is assumed

to hold for all integers k and n, and if (L) = 0 for all k < O?

35 Prove that, for every

> 0, there exists an integer n > 1 (depending

such that

mod 1 <

36 Is it possible to stack n bricks in such a way that the topmost brick is not

above any point of the bottommost brick, yet a person who weighs the

same as 100 bricks can balance on the middle of the top brick without

toppling the pile?

37 Express

I.,“=“,

(k mod

m)/k(k

+ 1) in terms of harmonic numbers, as-

suming that m and n are positive integers. What is the limiting value

asn-+co?

Ah! Those were

prime years.

38 Find the indefinite sum

(I)

(-l)kHk

6k.

39 Express

xz=,

Ht in terms of n and H,.

40 Prove that 1979 divides the numerator of

t~~,9(-l)k~‘/k,

and give a

similar result for 1987. Hint: Use Gauss’s trick to obtain a sum of

fractions whose numerators are 1979. See also exercise 4.

41 Evaluate the sum

in closed form, when n is an integer (possibly negative).

42 If S is a set of integers, let S + 1 be the “shifted” set {x + 1 1 x

S}.

How many subsets of

,2, . . , n} have the property that S U (S + 1) =

{1,2,...,n+l}?

43 Prove that the infinite sum

+.Ol

+.002

+.0003

+.00005

+.000008

+.0000013

converges to a rational number.

300 SPECIAL NUMBERS

44 Prove the converse of Cassini’s identity (6.106): If k and m are integers

such that Im2-km-k21 = 1, then there is an integer n such that k =

fF,

and m =

fF,+l.

45 Use the repertoire method to solve the general recurrence

= a;

x, = p;

= X,--l

+X,-2+yn+6.

What are cos 36” and cos 72”?

47 Show that

2"~'h =

;

(2;,)5k,

and use this identity to deduce the values of

mod p and

F,+1

mod p

when p is prime.

48 Prove that zero-valued parameters can be removed from continuant poly-

nomials by collapsing their neighbors together:

K,(xl,...

,xTl-1,0,x

m+l,...,Xn)

K,-2(x,,.

. . ,

Xm~Z,Xm~l+X,+l,X,+Z,...,X,),

l<m<n.

49 Find the continued fraction representation of the number

2-ln@J.

50 Define f(n) for all positive integers n by the recurrence

f(1) = 1;

f(2n) = f(n);

f(2nfl)

f(n)+f(n+l).

For which

f(n)

even?

b Show that f(n) can be expressed in terms of continuants.

Exam problems

51 Let p be a prime number.

a Prove that

{E}

[E]

0 (mod p), for 1 < k < p.

b Prove that

[“,‘I

1 (mod p), for 1

k < p.

Prove that {‘“;‘}

[‘“,-‘1

0 (mod p).

d Prove that if p > 3 we have [;] F 0 (mod

p2).

Hint: Consider

pp.

Let

be written in lowest terms as an/bn.

a Prove that p\b,,

+=+

p%aln,pJ,

if p is prime.

b Find all n > 0 such that a,, is divisible by 5.

6 EXERCISES 301

53 Find a closed form for

tkm,O

(E)-‘(-l)kHk,

when 0 6 m < n. Hint:

Exercise 5.42 has the sum without the Hk factor.

54 Let n > 0. The purpose of this exercise is to show that the denominator

Bz,,

is the product of all primes p such that (p-1)\(2n).

a Show that S,(p) + [(p-l)\

m is a multiple of p, when p is prime

]

and m > 0.

b Use the result of part (a) to show that

Bzn

[(p-‘)\(2n)l =

Izn

is an integer.

prime

Hint: It suffices to prove that, if p is any prime, the denominator of

the fraction

Bz,,

[(p-1)\(2n)]/p

is not divisible by

Prove that the denominator of

Bzn

is always an odd multiple of 6,

and it is equal to 6 for infinitely many n.

Prove (6.70) as a corollary of a more general identity, by summing

and differentiating with respect to x.

56 Evaluate

k+m

(;) t-1 lkkn+‘/(k-

m in closed form as a function of the

)

integers m and n. (The sum is over all integers k except for the value

k=m.)

The “wraparound binomial coefficients of order 5” are defined by

((;)>

((nk’))

((,k:;mod,))’

n>O’

and

((E))

= [k=Ol. Let

Q,,

be the difference between the largest and

smallest of these numbers in row n:

E5((L))

o%((;))

Find and prove a relation between

Q,,

and the Fibonacci numbers.

Find closed forms for

Fiz” and

tntO

F:zn.

What do you deduce

about the quantity Fi,,

4Fi

F:_,?

59 Prove that if m and n are positive integers, there exists an integer x such

that

m (mod

3”).

60 Find all positive integers n such that either F, + 1 or F,

1 is a prime

number.

302 SPECIAL NUMBERS

61 Prove the identity

integer n 3 1.

What is

~~=,

1 /FJ.2k?

62 Let A, =

4”

+ @-” and

4”

a-“.

Find constants

and

such that A,, =

aA,-1

@An-2

and

OLB~-I

+ BBn-2 for all n 3 0.

Express A,, and

in terms of

and

(see exercise 28).

Prove that

xE=,

1 ,/(Fzk+l + 1) =

B,/A,+l.

Find a closed form for

EL=,

l/(F~k+,

1).

Bonus

problems

Bogus problems

63 How many permutations

7~1~2..

. rrn of

{1,2,.

. . , n} have exactly k in-

dices j such that

rri <

7Cj

for all i < j? (Such j are called “left-to-right maxima!‘)

> j? (Such j are called “excedances!‘)

64 What is the denominator of

[,j/f,],

when this fraction is reduced to

lowest terms?

65 Prove the identity

n f(k)

. . .

0 0

f(lx,

+...+x,])dx,

..dx.

nl.

’

66 Show that ((y)) = 2(y), and find a closed form for ((y)).

67 Find a closed form for

Et=,

k’H,,+k.

68 Show that the generalized harmonic numbers of exercise 22 have the

power series expansion

H, =

x(-l)nHL)zn-‘.

n>2

69 Prove that the generalized factorial of equation (5.83) can be written

by considering the limit as n

of the first n factors of this infinite

product. Show that -&(z!) is related to the general harmonic numbers of

exercise 22. .

6 EXERCISES 303

70 Prove that the tangent function has the power series (6.g2), and find the

corresponding series for z/sin

and ln( (tan

2)/z).

71 Find a relation between the numbers

T,,

(1) and the coefficients of 1 /cos z.

72 What is

I.,(-l)“(L),

the row sum of Euler’s triangle with alternating

signs?

73 Prove that, for all integers n 3 1,

zcotz

4cot4--4tan-4_

2” 2” 2”

2"-1

k=l

cot

+cot

and show that the limit of the kth summand is 2z2/(z2

k2rr2)

for fixed k

as n

00.

74 Prove the following relation that connects Stirling numbers, Bernoulli

numbers, and Catalan numbers:

75 Show that the four chessboard pieces of the 64 = 65 paradox can also be

reassembled to prove that 64 = 63.

76 A sequence defined by the recurrence

A, ==x,

=y,

= An-1 + A,pz

has A,,, = 1000000 for some m. What positive integers x and y make m

as large as possible?

77 The text describes a way to change a formula involving

Fn*k

to a formula

that involves

and F,+j only. Therefore it’s natural to wonder if two

such “reduced” formulas can be equal when they aren’t identical in form.

Let P(x,y) be a polynomial in x and y with integer coefficients. Find a

necessary and sufficient condition that

P(F,+,

, F,) = 0 for all n 3 0.

78 Explain how to add positive integers, working entirely in the Fibonacci

number system.

79 Is it possible that a sequence (A,) satisfying the Fibonacci recurrence

A,, = A,-1 +

A,-2

can contain no prime numbers, if

and

are

relatively prime?

304 SPECIAL NUMBERS

80 Show that continuant polynomials appear in the matrix product

A)(;

J2)-.(Y

iI)

and in the determinant

det

-1

0 1 0 0 . . . 0 0

-1x31

0 -1

,..

-1 . . .

81 Generalizing (6.146), find a continued fraction related to the generating

function En21

z LnaJ,

when

is any positive irrational number.

82 Let m and n be odd, positive integers. Find closed forms for

F2,,*+:+F

;

Fzmk+:-Fm'

k>O

Hint: The sums in exercise 62 are S:,3

ST,,,,,

and

S1,s

ST,~,+~.

83 Let

be an irrational number in

(0,l)

and let al, a2,

as,

. . . be the

partial quotients in its continued fraction representation. Show that

(01,

n) 1 < 2 when n = K(

al,

. . . , a,), where D is the discrepancy

defined in Chapter 3.

84 Let Q,, be the largest denominator on level n of the Stern-Brocot tree.

(Thus

(Qo,

QI, Q2, Q3,Qh,. . .) =

(1,2,3,5,8,.

. .) according to the dia-

gram in Chapter 4.) Prove that Q,, = F,+2.

85 Characterize all N such that the Fibonacci residues

{FomodN,

modN,

FzmodN,

. ..}

form the complete set {0, 1,. . . , N

l}. (See exercise 59.)

Research problems

86 What is the best way to extend the definition of {t} to arbitrary real

values of n and k?

87 Let

be written in lowest terms as

an/b,,

as in exercise 52.

Are there infinitely many n with 11 \a,?

Are there infinitely many n with b, =

lcm(l,2,.

. .

,n)?

(Two such

values are n = 250 and n = 1000.)

88 Prove that y and

are irrational.

6 EXERCISES 305

89 Develop a general theory of the solutions to the two-parameter recurrence

(an+

@+y)

+(a’n+/3’k+y’)

+[n=k=OI,

forn,k30,

assuming that

[:I

= 0 h

en n < 0 or k < 0. (Binomial coefficients,

Stirling numbers, Eulerian numbers, and the sequences of exercises 17

and 31 are special cases.) What special values

(LX,

fl,r,

CX’,

fi’,~‘)

yield

“fundamental solutions” in terms of which the general solution can be

expressed?

Generating Functions

THE MOST POWERFUL WAY to deal with sequences of numbers, as far

as anybody knows, is to manipulate infinite series that “generate” those se-

quences.

We’ve learned a lot of sequences and we’ve seen a few generating

functions; now we’re ready to explore generating functions in depth, and to

see how remarkably useful they are.

7.1

DOMINO THEORY AND CHANGE

Generating functions are important enough, and for many of us new

enough, to justify a relaxed approach as we begin to look at them more closely.

So let’s start this chapter with some fun and games as we try to develop our

intuitions about generating functions. We will study two applications of the

ideas, one involving dominoes and the other involving coins.

How many ways

T,,

are there to completely cover a 2 x n rectangle with

2 x 1 dominoes? We assume that the dominoes are identical (either because

they’re face down, or because someone has rendered them indistinguishable,

say by painting them all red); thus only their orientations-vertical or hori-

zontal-matter, and we can imagine that we’re working with domino-shaped

tiles. For example, there are three tilings of a 2 x 3 rectangle, namely llll,

and Eli; so

= 3.

To find a closed form for general T, we do our usual first thing, look at

“Let me count the

small cases. When n = 1 there’s obviously just one tiling,

and when n = 2

ways.

”

there are two, •l and El.

-E. B. Browning

How about when n = 0; how many tilings of a 2 x 0 rectangle are there?

It’s not immediately clear what this question means, but we’ve seen similar

situations before: There is one permutation of zero objects (namely the empty

permutation), so O! =

There is one way to choose zero things from n things

(namely to choose nothing), so (t) = 1. There is one way to partition the

empty set into zero nonempty subsets, but there are no such ways to partition

a nonempty set; so

{:}

= [n =

01.

By such reasoning we can conclude that

306

7.1 DOMINO THEORY AND CHANGE 307

there’s just one way to tile a 2 x 0 rectangle with dominoes, namely to use

no dominoes; therefore

= 1. (This spoils the simple pattern

T,,

= n that

holds when n = 1, 2, and 3; but that pattern was probably doomed anyway,

since

wants to be 1 according to the logic of the situation.) A proper

understanding of the null case turns out to be useful whenever we want to

solve an enumeration problem.

Let’s look at one more small case, n = 4. There are two possibilities for

tiling the left edge of the rectangle-we put either a vertical domino or two

horizontal dominoes there. If we choose a vertical one, the partial solution is

and the remaining 2 x 3 rectangle can be covered in

ways. If we choose

two horizontals, the partial solution

can be completed in

ways. Thus

= 5. (The five tilings are

UIR,

UE, El, EII, and

M.)

We now know the first five values of

T,,:

These look suspiciously like the Fibonacci numbers, and it’s not hard to see

why: The reasoning we used to establish

easily generalizes to

T,,

= T,_l + Tn-2, for n > 2. Thus we have the same recurrence here as for

the Fibonacci numbers, except that the initial values

= 1 and

= 1 are a

little different. But these initial values are the consecutive Fibonacci numbers

and

F2,

so the T’s are just Fibonacci numbers shifted up one place:

F,+I

for n > 0.

(We consider this to be a closed form for

Tnr

because the Fibonacci numbers

are important enough to be considered “known!’ Also,

itself has a closed

form (6.123) in terms of algebraic operations.) Notice that this equation

confirms the wisdom of setting

= 1.

But what does all this have to do with generating functions? Well, we’re

about to get to that -there’s another way to figure out what

T,,

is. This new

‘lb boldly go

way is based on a bold idea. Let’s consider the “sum” of all possible 2 x n

where no tiling has

gone before.

tilings, for all n 3 0, and call it T:

=~+o+rn+~+m~+m+a+....

(7.1)

(The first term ‘I’ on the right stands for the null tiling of a 2 x 0 rectangle.)

This sum T represents lots of information. It’s useful because it lets us prove

things about T as a whole rather than forcing us to prove them (by induction)

about its individual terms.

The terms of this sum stand for tilings, which are combinatorial objects.

We won’t be fussy about what’s considered legal when infinitely many tilings

308 GENERATING FUNCTIONS

are added together; everything can be made rigorous, but our goal right now

is to expand our consciousness beyond conventional algebraic formulas.

We’ve added the patterns together, and we can also multiply them-by

juxtaposition. For example, we can multiply the tilings

and E to get the

new tiling iEi. But notice that multiplication is not commutative; that is, the

order of multiplication counts: [B is different from EL

Using this notion of multiplication it’s not hard to see that the null

tiling plays a special role--it is the multiplicative identity. For instance,

IxEi=Exl=E.

Now we can use domino arithmetic to manipulate the infinite sum T:

T = I+O+CI+E+Ull+CEl+Ell+~~~

~+o(~+o+m+8-t~~~)+8(~+0+m+e+~~~)

I+UT+HT.

(7.2)

Every valid tiling occurs exactly once in each right side, so what we’ve done is

reasonable even though we’re ignoring the cautions in Chapter 2 about “ab-

solute convergence!’ The bottom line of this equation tells us that everything

have a gut fee/-

in T is either the null tiling, or is a vertical tile followed by something else

ing that these

in T, or is two horizontal tiles followed by something else in T.

sums must con-

verge, as long as

So now let’s try to solve the equation for T. Replacing the T on the left

the dominoes are

by IT and subtracting the last two terms on the right from both sides of the sma”en’Ju&

equation, we get

(I-O-E)T = I.

(7.3)

For a consistency check, here’s an expanded version:

+...

-n-m-~-~-rJ-J-J-rjyg-rj=J

-...

-~-.a--EgJ-@=J-~-KJ-~

-...

Every term in the top row, except the first, is cancelled by a term in either

the second or third row, so our equation is correct.

So far it’s been fairly easy to make combinatorial sense of the equations

we’ve been working with. Now, however, to get a compact expression for T

we cross a combinatorial divide. With a leap of algebraic faith we divide both

sides of equation (7.3) by

I--O-E

to get

I-o-8’

(7.4)

7.1 DOMINO THEORY AND CHANGE 309

(Multiplication isn’t commutative, so we’re on the verge of cheating, by not

distinguishing between left and right division. In our application it doesn’t

matter, because I commutes with everything. But let’s not be picky, unless

our wild ideas lead to paradoxes.)

The next step is to expand this fraction as a power series, using the rule

1-z

1 +

+ . . . .

The null tiling I, which is the multiplicative identity for our combinatorial

arithmetic, plays the part of

the usual multiplicative identity; and

+ q

plays

So we get the expansion

I-U-El

I+I:o+E)+(u+E)2+(u+E)3+~~~

~+~:o+e)+(m+m+~+m)

+ (ml+uB+al+rm+Bn+BE+E3l+m3)

f...

This is T, but the tilings are arranged in a different order than we had before.

Every tiling appears exactly once in this sum; for example,

CEXE!ll

appears

in the expansion of

(

)‘.

We can get useful information from this infinite sum by compressing it

down, ignoring details that are not of interest. For example, we can imagine

that the patterns become unglued and that the individual dominoes commute

with each other; then a term like IEEIB becomes

C1406,

because it contains

four verticals and six horizontals. Collecting like terms gives us the series

=I+O+02-to2+03+2002t04+30202+~4+~~~.

The

here represents the two terms of the old expansion,

and

ELI,

that

have one vertical and two horizontal dominoes; similarly

302

0’

represents the

three terms CB, CH, and Elll. We’re essentially treating

and o as ordinary

(commutative) variables.

We can find a closed form for the coefficients in the commutative version

of T by using the binomial theorem:

I-

+ 02)

= I+(o+o~)+(o+,~)~+(o+~~)~+...

~(Ofo2)k

k>O

(7d

310 GENERATING FUNCTIONS

(The last step replaces k-j by m; this is legal because we have (1) = 0 when

0 6 k < j.) We conclude that

(‘;“)

is the number of ways to tile a 2 x (j

+2m)

rectangle with j vertical dominoes and 2m horizontal dominoes. For example,

we recently looked at the 2 x 10 tiling CERIRJ, which involves four verticals

and six horizontals; there are (“1”) = 35 such tilings in all, so one of the terms

in the commutative version of T is

350406.

We can suppress even more detail by ignoring the orientation of the

dominoes. Suppose we don’t care about the horizontal/vertical breakdown;

we only want to know about the total number of 2 x n tilings. (This, in

fact, is the number T, we started out trying to discover.) We can collect

the necessary information by simply substituting

single quantity,

for

and O. And we might as well also replace I by 1, getting

Now

I’m

dis-

oriented.

l-z-22'

(7.6)

This is the generating function (6.117) for Fibonacci numbers, except for a

missing factor of

in the numerator; so we conclude that the coefficient of

Z”

in T is F,+r .

The compact representations

I/(1-O-R),

I/(I-O-EI~),

and 1/(1-z-z')

that we have deduced for

are called generating functions, because they

generate the coefficients of interest.

Incidentally, our derivation implies that the number of 2 x n domino

tilings with exactly m pairs of horizontal dominoes is (“-,“). (This follows

because there are j = n

2m vertical dominoes, hence there are

(i:m)

(j+J

(“m”)

ways to do the tiling according to our formula.) We observed in Chapter 6

that (“km) is the number of Morse code sequences of length n that contain

m dashes; in fact, it’s easy to see that 2 x n domino tilings correspond directly

to Morse code sequences. l(The tiling

CEEURI

corresponds to

‘a-

-*a -*‘.)

Thus domino tilings are closely related to the continuant polynomials we

studied in Chapter 6. It’s a small world.

We have solved the

problem in two ways. The first way, guessing the

answer and proving it by induction, was easier; the second way, using infinite

sums of domino patterns and distilling out the coefficients of interest, was

fancier. But did we use the second method only because it was amusing to

play with dominoes as if they were algebraic variables? No; the real reason

for introducing the second way was that the infinite-sum approach is a lot

more powerful. The second method applies to many more problems, because,

it doesn’t require us to make magic guesses.

7.1 DOMINO THEORY AND CHANGE 311

Let’s generalize up a notch, to a problem where guesswork will be beyond

us. How many ways Ll, are there to tile a 3 x n rectangle with dominoes?

The first few cases of this problem tell us a little: The null tiling gives

There is no valid tiling when n = 1, since a 2 x 1 domino doesn’t fill

a 3 x 1 rectangle, and since there isn’t room for two. The next case, n = 2,

can easily be done by hand; there are three tilings,

m, and R, so

= 3.

(Come to think of it we already knew this, because the previous problem told

us that

= 3; the number of ways to tile a 3 x 2 rectangle is the same as the

number to tile a 2 x 3.) When n = 3, as when n =

there are no tilings. We

can convince ourselves of this either by making a quick exhaustive search or

by looking at the problem from a higher level: The area of a 3 x 3 rectangle is

odd, so we can’t possibly tile it with dominoes whose area is even. (The same

argument obviously applies to any odd n.) Finally, when n = 4 there seem

to be about a dozen tilings; it’s difficult to be sure about the exact number

without spending a lot of time to guarantee that the list is complete.

So let’s try the infinite-sum approach that worked last time:

=I+E9+f13+~+W+~-tW+e4+~+....

(7.7)

Every non-null tiling begins with either 0 or

but unfortunately the

first two of these three possibilities don’t simply factor out and leave us with

again. The sum of all terms in

that begin with 0 can, however, be written

LV,

where

=~+g+~+g+Q+...

is the sum of all domino tilings of a mutilated 3 x n rectangle that has its

lower left corner missing. Similarly, the terms of

that begin with

Ei’

can be

written FA, where

consists of all rectangular tilings lacking their upper left corner. The series A

is a mirror image of V. These factorizations allow us to write

= I

+0V+-BA+pJl.

And we can factor V and A as well, because such tilings can begin in only

two ways:

v =

ml+%V,

A =

gU+@A.

312 GENERATING FUNCTIONS

Now we have three equations in three unknowns (U, V, and A). We can solve

them by first solving for V and A in terms of U, then plugging the results

into the equation for U:

Q)-ml,

(I-g)-‘ou;

B(l-B,)-‘ml

B(I-

gyou

pJu

And the final equation can be solved for U, giving the compact formula

B(l-@)-‘[I

B(I-gJ-‘o

R’

(7.8)

This expression defines the infinite sum U, just as (7.4) defines T.

The next step is to go commutative.

Everything simplifies beautifully

when we detach all the dominoes and use only powers of

and =:

O&(1

,3)-~’

Po(l

,3)-l

l-o3

(I-

,3)2

-20%;

c33)-’

l-202

o(1

&:I+

2020

404 02

80603

=m+

,3)3

,3)5

,3)7

+...

(m;2k)2’.,,2kak+h.

k,m>O

(This derivation deserves careful scrutiny. The last step uses the formula

,)-2k--1 =

(m+mZk)Wm,

identity (5.56).) Let’s take a good look at

the bottom line to see what it tells us. First, it says that every 3 x n tiling

uses an even number of vertical dominoes. Moreover, if there are 2k verticals,

there must be at least k horizontals, and the total number of horizontals must

be k + 3m for some m 3 0. Finally, the number of possible tilings with 2k

verticals and k + 3m horizontals is exactly (“i2k)2k.

We now are able to analyze the 3 x 4 tilings that left us doubtful when we

began looking at the 3 x n problem. When n = 4 the total area is 12, so we

need six dominoes altogether. There are 2k verticals and k + 3m horizontals,

I /earned in another

class about “regular

expressions.” If

I’m

not mistaken, we

can write

u =

(LB,*0

+BR*o+H)*

in the language of

regular expressions;

so there must be

some connection

between regular

expressions and gen-

erating functions.

7.1 DOMINO THEORY AND CHANGE 313

for some k and

hence 2k + k + 3m = 6. In other words, k + m = 2.

If we use no

vertic:als,

then k = 0 and m = 2; the number of possibilities

is (Zt0)20 = 1. (This accounts for the tiling

B.)

If we use two verticals,

then k = 1 and m = 1; there are (‘t2)2’ = 6 such tilings. And if we use

four verticals, then k = 2 and m = 0; there are (“i4)22 = 4 such tilings,

making a total of

114

= 11. In general if n is even, this reasoning shows that

k + m =

in,

hence

(mL2k)

($5’:)

and the total number of 3 x n tilings is

(7.9)

As before, we can also substitute z for both

and O, getting a gen-

erating function that doesn’t discriminate between dominoes of particular

persuasions. The result is

u=-

-z3

-z3(1

-9-l

-z3(1

-9-1

-z3

l-423

$26.

(7.10)

If we expand this quotient into a power series, we get

U = 1

+U2z”+U4Z6+U~Z9+UsZ12+~~~,

a generating function for the numbers U,. (There’s a curious mismatch be-

tween subscripts and exponents in this formula, but it is easily explained. The

coefficient of z9, for example, is

Ug,

which counts the tilings of a 3 x 6 rectan-

gle. This is what we want, because every such tiling contains nine dominoes.)

We could proceed to analyze (7.10) and get a closed form for the coeffi-

cients, but it’s

bett,er

to save that for later in the chapter after we’ve gotten

more experience. So let’s divest ourselves of dominoes for the moment and

proceed to the next advertised problem, “change!’

How many ways are there to pay 50 cents? We assume that the payment

must be made with pennies

nickels

dimes

quarters

and half-

Ah yes, I remember

dollars

George Polya

[239]

popularized this problem by showing that it

when we had

half-

dollars.

can be solved with generating functions in an instructive way.

Let’s set up infinite sums that represent all possible ways to give change,

just as we tackled the domino problems by working with infinite sums that

represent all possible domino patterns. It’s simplest to start by working with

fewer varieties of coins, so let’s suppose first that we have nothing but pennies.

The sum of all ways to leave some number of pennies (but just pennies) in

change can be written

%+o+oo+ooo+oooo+

J+O+02+03+04+...

314 GENERATING FUNCTIONS

The first term stands for the way to leave no pennies, the second term stands

for one penny, then two pennies, three pennies, and so on. Now if we’re

allowed to use both pennies and nickels, the sum of all possible ways is

since each payment has a certain number of nickels chosen from the first

factor and a certain number of pennies chosen from P. (Notice that N is

not the sum { +

+ O)2 +

@)3

+ . . . , because such a

sum includes many types of payment more than once. For example, the term

+ @)2 =

treats

and

as if they were

different, but we want to list each set of coins only once without respect to

order.)

Similarly, if dimes are permitted as well, we get the infinite sum

D =

(++@+@2+@3+@4+..)N,

which includes terms like

@3@3@5

= @@@@@@@@O@@ when it is

expanded in full. Each of these terms is a different way to make change.

Adding quarters and then half-dollars to the realm of possibilities gives

Coins of the realm.

Q =

(++@+@2+@3+@4+...)D;

C =

(++@+@2+@3+@4+-.)Q.

Our problem is to find the number of terms in C worth exactly

509!.

A simple trick solves this problem nicely: We can replace

by z,

by z5,

z”,

@ by

z25,

and

z50.

Then each term is replaced by

zn,

where n is the monetary value of the original term. For example, the term

@@@@@ becomes

z50+10f5+5+’

2”.

The four ways of paying 13 cents,

namely @,03, @OS,

0203,

and 013, each reduce to

z13;

hence the coefficient

z13

will be 4 after the z-substitutions are made.

Let

P,,

N,,

D,,

Qn,

and

be the numbers of ways to pay n cents

when we’re allowed to use coins that are worth at most

5, 10, 25, and 50

cents, respectively. Our analysis tells us that these are the coefficients of 2”

in the respective power series

P = 1 + z +

+ . .

)

N = (1

+~~+z’~+z’~‘+z~~+...)P,

D =

(1+z’0+z20+z”0+z40+...)N,

Q = (1

+z25+z50+z;‘5+~‘oo+~~~)D,

+,50+z’00+z’50+Z200+...)Q~

7.1 DOMINO THEORY AND CHANGE 315

How many pennies

Obviously P, = 1 for all n 3 0. And a little thought proves that we have

are

there, really?

If n is greater

= Ln/5J + 1: To make n cents out of pennies and nickels, we must choose

than, say,

10”)

either 0 or

or . . . or

Ln/5] nickels, after which there’s only one way to supply

bet that

= 0

the requisite number of pennies. Thus P, and

are simple; but the values

in the “real world.”

Dn,

Qn,

and

are increasingly more complicated.

One way to deal with these formulas is to realize that 1 +

+ 2’“’

. .

is just

l/(1

2”‘). Thus we can write

P =

l/(1

-2’1,

N = P/(1

-i’),

D =

N/(1

2”)

Q =

D/(1

zz5)

C = Q/(1

-2”).

Multiplying by the denominators, we have

(l-z)P

= 1,

-z5)N

= P,

(l-z”)D

= N,

(~-z~~)Q

= D,

(1-z5’)C

= Q.

Now we can equate coefficients of 2” in these equations, getting recurrence

relations from which the desired coefficients can quickly be computed:

P,-I

+ [n=O] ,

N-5

P,,

D, =

Dn-IO

-tN,,

Qn =

Qn-25

-t

D,,

G-50

+ Qn.

For example, the coefficient of

Z”

in D = (1

z~~)Q is equal to Q,,

Qnp25;

so we must have Qll

Qnp25 = D,, as claimed.

We could unfold these recurrences and find, for example, that Qn =

D,+D,-zs+Dn~5o+Dn~75+...,

stopping when the subscripts get negative.

But the non-iterated form is convenient because each coefficient is computed

with just one addition, as in Pascal’s triangle.

Let’s use the recurrences to find

Csc.

First,

Cso

Q50;

so we want

to know Qso. Then

Q50

Q25

D50,

and

Q25

+ D25; so we also want

to know

D50

and 1125. These

depend in turn on

DUO,

D15,

DIO,

D5,

and

NSO,

NC,,

Ns. A simple calculation therefore suffices to

316

GENERATING FUNCTIONS

determine all the necessary coefficients:

n 0 5

10 15 20 25

30 35 40 45 50

11111111111

NTI

12345 6 7

9 10 11

12 4 6 9 1216 25 36

13 49

The final value in the table gives us our answer,

COO:

There are exactly 50 ways

to leave a 50-cent tip.

(Not counting the

How about a closed form for C,? Multiplying the equations together

Option

ofchar@ng

gives us the compact expression

the tip to a

credit

card.)

1 1 1

----~~

--z

--5

-zz~o

-z25

-z50

(7.11)

but it’s not obvious how to get from here to the coefficient of

zn.

Fortunately

there is a way; we’ll return to this problem later in the chapter.

More elegant formulas arise if we consider the problem of giving change

when we live in a land that mints coins of every positive integer denomination

(0,

. . .

) instead of just the five we allowed before. The corresponding

generating function is an infinite product of fractions,

(1 -z)(l -22)(1 -23)..1'

and the coefficient of 2” when these factors are fully multiplied out is called

p(n), the number of partitions of n. A partition of n is a representation of n

as a sum of positive integers, disregarding order. For example, there are seven

different partitions of 5, namely

5=4+1=3+2=3+11-1=2+2+1=2+1+1+1=1+1+1+1+1;

hence p(5) = 7. (Also p(2)

2, p(3) = 3, p(4) = 5, and p(6) = 11; it begins

to look as if p(n) is always a prime number. But

7) = 15, spoiling the

pattern.) There is no closed form for p(n), but the theory of partitions is a

fascinating branch of mathematics in which many remarkable discoveries have

been made. For example, Ramanujan proved that p(5n + 4) E 0 (mod 5),

p(7n + 5)

0 (mod 7), and

p(1

+ 6) E 0 (mod 1

l),

by making ingenious

transformations of generating functions (see Andrews

[ll,

Chapter

lo]).

physicists

can

get

away with viewing

light sometimes as

a wave and some-

times as a particle,

mathematicians

should be able to

view generating

functions in two

different ways.

7.2 BASIC MANEUVERS 317

7.2 BASIC MANEUVERS

Now let’s look more closely at some of the techniques that make

power series powerful.

First a few words about terminology and notation. Our generic generat-

ing function has the form

G(z) =

go+glz+gzz’+-.

xg,,z”,

(7.12)

n>o

and we say that G(z), or G for short, is the generating function for the se-

quence

(m,gl,a,...),

h’

w ic we also call (gn). The coefficient

g,,

of zn

in G(z) is sometimes denoted

[z”]

G(z).

The sum in

(7.12)

runs over all n 3 0, but we often find it more con-

venient to extend the sum over all integers n. We can do this by simply

regarding g-1 = g-2 =

...

= 0. In such cases we might still talk about the

sequence

(90,91,92,.

. . ),

as if the g,‘s didn’t exist for negative n.

Two kinds of “closed forms” come up when we work with generating

functions. We might have a closed form for G(z), expressed in terms of z; or

we might have a closed form for

gnr

expressed in terms of n. For example, the

generating function for Fibonacci numbers has the closed form z/( 1

z2);

the Fibonacci numbers themselves have the closed form

(4”

$n)/fi.

The

context will explain what kind of closed form is meant.

Now a few words about perspective. The generating function G(z) ap-

pears to be two different entities, depending on how we view it. Sometimes

it is a function of a complex variable z, satisfying all the standard properties

proved in calculus books. And sometimes it is simply a formal power series,

with z acting as a placeholder. In the previous section, for example, we used

the second interpretation; we saw several examples in which z was substi-

tuted for some feature of a combinatorial object in a “sum” of such objects.

The coefficient of

Z”

was then the number of combinatorial objects having n

occurrences of that feature.

When we view G(z) as a function of a complex variable, its convergence

becomes an issue. We said in Chapter 2 that the infinite series

gnzn

converges (absolutely) if and only if there’s a bounding constant A such that

the finite sums

O.SnSN

/gnznl

never exceed A, for any N. Therefore it’s easy

to see that if

tn3c

gnzn converges for some value z =

it also converges for

all z with

IzI

1~01.

Furthermore, we must have

lim,,,

lgnzzl = 0; hence, in

the notation of Chapter 9,

O(ll/z#)

if there is convergence at

~0.

And

conversely if

= O(Mn), the series

nao

gnzn converges for all

IzI

< l/M.

These are the basic facts about convergence of power series.

But for our purposes convergence is usually a red herring, unless we’re

trying to study the asymptotic behavior of the coefficients. Nearly every

318 GENERATING FUNCTIONS

operation we perform on generating functions can be justified rigorously as

an operation on formal power series, and such operations are legal even when

the series don’t converge. (The relevant theory can be found, for example, in

Bell

[19],

Niven

[225],

and Henrici [151, Chapter

11.)

Furthermore, even if we throw all caution to the winds and derive formu-

Even if we remove

las without any rigorous justification, we generally can take the results of our

the

ta@

frem

Our

derivation and prove them by induction. For example, the generating

func-

mat tresses.

tion for the Fibonacci numbers converges only when

/zI

l/4

0.618, but

we didn’t need to know that when we proved the formula

(4”

Gn)/&.

The latter formula, once discovered, can be verified directly, if we don’t trust

the theory of formal power series. Therefore we’ll ignore questions of conver-

gence in this chapter; it’s more a hindrance than a help.

So much for perspective. Next we look at our main tools for reshaping

generating functions-adding, shifting, changing variables, differentiating,

integrating, and multiplying. In what follows we assume that, unless stated

otherwise, F(z) and G(z) are the generating functions for the sequences (fn)

and (gn). We also assume that the f,,‘s and g,‘s are zero for negative n, since

this saves us some bickering with the limits of summation.

It’s pretty obvious what happens when we add constant multiples of

F and G together:

aF(z)

BG(z) =

atf,,z”

gnzn

trf,+

fig,)?.

(7.13)

This gives us the generating function for the sequence (af, + Bgn).

Shifting a generating function isn’t much harder. To shift G(z) right by

m places, that is, to form the generating function for the sequence (0,. . .

,O,

90,91,...

) = (gnPm) with m. leading O’s, we simply multiply by zm:

zmG(z) =

g,,

z”+“’

g+,,,z”,

integer m 3 0.

(7.14)

n n

This is the operation we used (twice), along with addition, to deduce the

equation (1

z’)F(z) =

on our way to finding a closed form for the

Fibonacci numbers in Chapter 6.

And to shift G(z) left m places-that is, to form the generating function

for

the

sequence

(sm, a,,+], gm+2,.

. . ) = (gn+,,,) with the first m elements

discarded- we subtract off the first m terms and then divide by

G(z)-go-g,z-. . .

-g,-,zm-l

gnPrn

h+mZ

(7.15)

n>m

ll>O

(We can’t extend this last sum over all n unless

= . . . =

gmPl

= 0.)

7.2 BASIC MANEUVERS 319

Replacing the

by a constant multiple is another of our tricks:

G(u)

~,(cz)~ =

xcngnz”;

(7.16)

this yields the generating function for the sequence

(c”g,).

The special case

c = -1 is particularly useful.

I fear d genera

ting-

Often we want to bring down a factor of n into the coefficient. Differen-

function dz

tiation is what lets us ‘do that:

G’(z) =

+2g2z+3g3z2+-

t(n+l)g,+,z".

(7.17)

Shifting this right one place gives us a form that’s sometimes more useful,

zG’(z)

tng,,z”

(7.18)

This is the generating function for the sequence (ng,). Repeated differentia-

tion would allow us to multiply

g,,

by any desired polynomial in n.

Integration, the inverse operation, lets us divide the terms by n:

G(t)dt

gez+

fg,z2

+ ;g2z3

+...

p-d.

(7.19)

TI>l

(Notice that the constant term is zero.) If we want the generating function

for

(g,/n)

instead of

(g+l/n),

we should first shift left one place, replacing

G(t) by (G(t)

gc)/t

in the integral.

Finally, here’s how we multiply generating functions together:

F(z)G(z)

(fo+f,z+f2z2+~-)(go+g1z+g2z2+-~)

(fogo) + (fog1 +f1!Ilo)z

(fog2

+f1g1

+f2go)z2

...

~(-pk&k)ZTI.

(7.20)

As we observed in Chapter 5, this gives the generating function for the se-

quence (hn), the convolution of (fn) and (gn). The sum

fk&-k

can

also be written h, =

~~=,

fkgnpkr because fk = 0 when k < 0 and gn-k = 0

when k > n. Multiplication/convolution is a little more complicated than

the other operations, but it’s very useful-so useful that we will spend all of

Section 7.5 below looking at examples of it.

Multiplication has several special cases that are worth considering as

operations in themselves. We’ve already seen one of these: When F(z) = z”’

we get the shifting operation (7.14). In that case the sum h,, becomes the

single term gnPm, because

all

fk's

0 except for

320 GENERATING FUNCTIONS

Table 320 Generating function manipulations.

aF(z)

K(z)

t(h

Bsn)z”

PG(z) =

gn-mz

integer m 3 0

G(~)-go-g,z-...-g,~,z~~’

;;

gn+mz ,

integer m 3 0

n20

G(a) =

~cngnzn

G’(z)

x(n+

l)gn+l

zG’(z) =

xngnz”

G(t) dt =

;gn.-,

2”

lI>l

F(z)G(z)

t(tfxg,,)z”

+;W

;(;g+

kin

Another useful special case arises when F(z) is the familiar function

1/(1--z)

1+z+z2+...;

then all

fk's

(for k 3 0) are 1 and we have

the important formula

&(z)

@<h-k)~n

t(tgk)z".

(7.21)

k>O

k<n

Multiplying a generating function by

l/(

l-z) gives us the generating function

for the cumulative sums of the original sequence.

Table 320 summarizes the operations we’ve discussed so far. To use

all these manipulations effectively it helps to have a healthy repertoire of

generating functions in stock. Table 321 lists the simplest ones; we can use

those to get started and to solve quite a few problems.

Each of the generating functions in Table 321 is important enough to

be memorized. Many of them are special cases of the others, and many of

7.2 BASIC MANEUVERS 321

Hint: 1f the se-

quence consists

of binomial coefi-

cients,

its generat-

ing function usually

involves a binomial,

1+z.

Table 321 Simple sequences and their generating functions.

sequence

generating function closed form

, o,o,

o,o,. .

)

(0,. . .

O,l,O,O

,...

(l,l,l,l,l,l,...)

(1,-1,1,-1,1,-l,...)

(l,O, l,O, l,O,. . .

)

(1,0,...,0,1,0,....0,1,0,

(1,43,4,5,6,...)

(1,2,4,8,16,32,...)

(1,4,6,4,1,0,0,...)

(k(;),(;),...)

(Lc,(':'),(':')

,...)

(l,c,cQ3,...)

(1,

(mm+'),

(mm+2),

("Z3),

(o,L;>;,$,...)

(OJ-;,;,-;,...)

(

11'111

)

‘2’6’24’,20””

,>o[n=Ol

fIoLn=ml

’

n30

tn>Op

1”

tn>O

[AnI

)

tn>O

[m\nlC

xn>o

n>O

2”

xn:O

(

)

t..-.(

)

EnI

(":"j

n>O

Loi

m+n

iz:

)

n2l n

(-v+’

n31

7x20

1-Z

l+z

l-22

l-zm

2)2

l-22

2J4

(1 +

z)C

l-cz

z)m+'

1-Z

ln(1 +

them can be derived quickly from the others by using the basic operations of

Table 320; therefore the memory work isn’t very hard.

For example, let’s consider the sequence

(1,2,3,4,

. ), whose generating

function

l/(

z)~

is often useful. This generating function appears near the

322

GENERATING FUNCTIONS

middle of Table 321, and it’s also the special case m = 1 of (1,

(","),

(mzL),

(“,‘“), ), which appears further down; it’s also the special case c = 2 of

the closely related sequence (1, c,

(‘:‘)

(‘12), .

We can derive it from the

generating function for (1 , 1 , 1 ,

. .

)

by taking cumulative sums as in (7.21);

that is, by dividing 1 /(l-z) by

-z).

Or we can derive it from

(1 , 1 , 1 ,

. ) OK, OK,

I’m

con-

by differentiation, using (7.17).

vinced

already

The sequence (1 , 0, 1 , 0, . ) is another one whose generating function can

be obtained in many ways. We can obviously derive the formula

zZn

l/(

z2) by substituting

for

in the identity

Z”

l/(

z);

we can

also apply cumulative summation to the sequence

(1, -1 ,

-1, . . . ),

whose

generating function is

l/(1

$ z), getting

l/(1

+z)(l

z) =

l/(1

-2’).

And

there’s also a third way, which is based on a general method for extracting

the even-numbered terms

(gc

, 0,

g2,

0, g4,0, . . . ) of any given sequence: If we

add

G(-z)

G(+z)

we get

G(Z)+

G(-z)

gn(l

+(-1)")~" =

g,[n

evenlz”;

therefore

G(z)

G(-z)

g2n

zLn

The odd-numbered terms can be extracted in a similar way,

G(z)

G(-z)

g2n+1zZn+'

(7.22)

In the special case where

g,,

1 and G(z) =

l/(

-z),

the generating function

for(1,0,1,0,...)is~(~(z)+~(-z))=t(&+&)=A.

Let’s try this extraction trick on the generating function for Fibonacci

numbers. We know that

I.,

F,zn =

z/(

2');

hence

F2nz

;(j57+l+r’,)

(

z3 z2

-z2)2-22

)

l-322+24

This generates the sequence

(Fo,

F2,0,

F4,.

. .

);

hence the sequence of alter-

nate F’s,

(Fo,Fl,Fd,F6,...)

(0,1,3,8,...

has a simple generating function:

F2,,zn

l-3z+z2

(7.24)

7.3 SOLVING RECURRENCES 323

7.3 SOLVING RECURRENCES

Now let’s focus our attention on one of the most important uses of

generating

functiorrs:

the solution of recurrence relations.

Given a sequence

(gn)

that satisfies a given recurrence, we seek a closed

form for

in terms of n. A solution to this problem via generating functions

proceeds in four steps that are almost mechanical enough to be programmed

on a computer:

Write down a single equation that expresses

g,,

in terms of other elements

of the sequence. This equation should be valid for all integers n, assuming

that g-1 = g-2 =

...

= 0.

2 Multiply both sides of the equation by zn and sum over all

This gives,

on the left, the sum

x.,

gnzn, which is the generating function G (2). The

right-hand side should be manipulated so that it becomes some other

expression involving G (2).

Solve the resulting equation, getting a closed form for G (2).

Expand G(z) into a power series and read off the coefficient of

zn;

this is

a closed form for

gn.

This method works because the single function G(z) represents the entire

sequence

(gn)

in such a way that many manipulations are possible.

Example 1: Fibonacci numbers revisited.

For example, let’s rerun the derivation of Fibonacci numbers from Chap-

ter 6. In that chapter we were feeling our way, learning a new method; now

we can be more systematic. The given recurrence is

go =

91 =

%-1+%-z,

for n 3 2.

We will find a closed form for

g,,

by using the four steps above.

Step 1 tells us to write the recurrence as a “single equation” for

gn.

could say

ifn<O;

if n = 1;

gn-1 -t gn-2, if n > 1;

but this is cheating. Step 1 really asks for a formula that doesn’t involve a

case-by-case construction. The single equation

gn-l+~ln-z

works for n > 2,

a.nd

it also holds when n 6 0 (because we have

= 0

and

gnegative

0). But when n = 1 we get 1 on the left and 0 on the right.

324 GENERATING FUNCTIONS

Fortunately the problem is easy to fix, since we can add [n =

to the right;

this adds 1 when n =

and it makes no change when n # 1. So, we have

s-1

+a-2+[n=ll;

this is the equation called fo:r in Step 1.

Step 2 now asks us to t:ransform the equation for (g,,) into an equation

for G(z) =

gnzn.

The task is not difficult:

G(z)

gnzn

~gnlzn+tg,~rzn+~[n=l]zn

;gnzn+l+;gnzn+2

fnz

G(z)

z’G(z)

+ z.

Step 3 is also simple in this case; we have

G(z) =

l-z-z2'

which of course comes as no surprise.

Step 4 is the clincher. We carried it out in Chapter 6 by having a sudden

flash of inspiration; let’s go more slowly now, so that we can get through

Step 4 safely later, when we meet problems that are more difficult. What is

b”l

l-z-22'

the coefficient of zn when z/( 1

z2) is expanded in a power series? More

generally, if we are given any rational function

P(z)

R(z) =

Qo,

where P and Q are polynomials, what is the coefficient

[z”]

R(z)?

There’s one kind of rational function whose coefficients are particularly

nice, namely

puz)m+1

(m;n)ap"z"

n30

(7.25)

(The case

= 1 appears in Table 321, and we can get the general formula

shown here by substituting

for z.) A finite sum of functions like (7.25),

s(z) =

pyl,-,+,

-p2Z)m2+'

+'.'+

-pLZ)mL+l

(7.4

7.3 SOLVING RECURRENCES 325

also has nice coefficients,

+ . . . + al

(7.27)

We will show that every rational function R(z) such that R(0) #

can be

expressed in the form

R(z)

S(z)

T(z),

(7.28)

where S(z) has the form (7.26) and T(z) is a polynomial. Therefore there is a

closed form for the coefficients

[z”]

R(z). Finding S(z) and T(z) is equivalent

to finding the “partial fraction expansion” of R(z).

Notice that S(z) =

when

has the values

l/p,,

. . . ,

l/pi.

Therefore

the numbers

that we need to find, if we’re going to succeed in expressing

R(z) in the desired form S(z) + T(z), must be the reciprocals of the numbers

where Q(ak) = 0. (Recall that R(z) =

P(z)/Q(z),

where P and Q are

polynomials; we have R(z) =

only if Q(z) = 0.)

Suppose Q(z) has the form

Q(z)

qo+q1z+~~~+q,z”‘,

where

0 and

q,,,

# 0.

The “reflected” polynomial

QR(z)

qoP+

q,z"-'

+...f

q,,,

has an important relation to Q (2):

QR(4 = qo(z

. .

P,)

Q(z) =

qo(l

-PIZ)...(~

-P~z)

Thus, the roots of QR are the reciprocals of the roots of Q, and vice versa.

We can therefore find the numbers

we seek by factoring the reflected poly-

nomial QR(z).

For example, in the Fibonacci case we have

Q(z) = 1

-2-z’;

QR(z) =

z2-z-l.

The roots of QR ca.n be found by setting (a, b, c) = (1, -1, -1) in the quad-

ratic formula (-b

II:

da)/2a;

we find that they are

l+ds

+=2

1-d

and $ =

Therefore QR(z) =

(z-+)(2-$)

and Q(z) = (1 -+z)(l

-i$z).

326 GENERATING FUNCTIONS

Once we’ve found the p’s, we can proceed to find the partial fraction

expansion. It’s simplest if all the roots are distinct, so let’s consider that

special case first. We might

a.s

well state and prove the general result formally:

Rational Expansion Theorem for Distinct Roots.

R(z)

= P(z)/Q(z), where Q(z) = qo(l

plz)

. . . (1

pLz)

and the

numbers (PI, . . . ,

PL)

are distinct, and if P(z) is a polynomial of degree less

than

then

[z”IR(z) =

a,p;+..+alp:,

-pkp(l/pk)

where

Q,fl,Pkl

(7.29)

Proof: Let

al,

. , . ,

be the stated constants. Formula (7.29) holds if R(z) =

P(z)/Q(z) is equal to

S(z)

d!-

-P1Z

+...+al.

PLZ

And we can prove that R(z) = S(z) by showing that the function T(z) =

R(z)

S(z) is not infinite as z

1 /ok. For this will show that the rational

Impress your par-

function T(z) is never infinite; hence T(z) must be a polynomial. We also can

ents

leaving

the

show that T(z)

0 as z

co; hence T(z) must be zero.

book open at this

page.

Let ak =

l/pk.

To prove that

lim,,,,

T(z) #

oo,

it suffices to show that

lim,,.,

cck)T(z)

= 0, because T(z) is a rational function of z. Thus we

want to show that

lim (Z

ak)R(Z)

;jzk

xk)s(z)

L’CCI,

The right-hand limit equals

l.im,,,,

ok(z-

c&)/‘(l

pkz) =

-ak/pk,

because

pkz) =

-pk(z-Kk)

and

(z-c&)/(1

PjZ)

-+ 0 for

# k. The left-hand

limit is

by L’Hospital’s rule. Thus the theorem is proved.

Returning to the Fibonacci example, we have P(z) = z and Q(z) =

= (1

@z)(l

$2); hence Q’(z) = -1

22, and

-PP(l/P)

-1

Q/(1/p)

-1 -2/p

=p+2.

According to (7.2g), the coefficient of +” in

[zn]

R(z) is therefore

@/(c$

+ 2) =

l/d;

the coefficient of

$”

$/($

+ 2) =

-l/\/5.

So the theorem tells us

that

= (+”

$“)/fi,

as in (6.123).

7.3 SOLVING RECURRENCES 327

When Q(z) has repeated roots, the calculations become more difficult,

but we can beef up the proof of the theorem and prove the following more

general result:

General Expansion Theorem for Rational Generating Functions.

R(z)

= P(t)/Q(z), where Q(z) =

qo(1

~12)~'

..(l

p~z)~[

and the

numbers

(PI,.

. , pi) are distinct, and if

P(z)

is a polynomial of degree less

than

+ . . . + dl, then

[z"]

R(z)

f,ln)p;

...

ft(n)p;

for all n 3 0,

(7.30)

where each fk(n) is a polynomial of degree

1 with leading coefficient

(7.31)

This can be proved by induction on max(dl , . . . , dl), using the fact that

al(dl

-l)!

R(z)

(1py

. . .

al(dl

l)!

WldL

is a rational function whose denominator polynomial is not divisible by

pkz)dk

for

any k.

Example 2: A more-or-less random recurrence.

Now that we’ve seen some general methods, we’re ready to tackle new

problems. Let’s try to find a closed form for the recurrence

go =

= 1 ;

gn-l+2g,~~+(-l)~,

for n 3 2.

(7.32)

It’s always a good idea to make a table of small cases first, and the recurrence

lets us do that easily:

No closed form is evident, and this sequence isn’t even listed in Sloane’s

Handbook

[270];

so we need to go through the four-step process if we want

to discover the solution.

328 GENERATING FUNCTIONS

Step 1 is easy, since we merely need to insert fudge factors to fix things

when n < 2: The equation

C.h-1

+&h-2

I-l)“[n~O]

[n=l]

holds for all integers n. Now we can carry out Step 2:

G(z)

g,,z”

gn-1zn+

gn-2zn

+t(-l)v+

n&l

n=l

N.B.: The upper

A(z)

2z2G(z) +

index on

En=,

z”

is not missing!

(Incidentally, we could also have used

(-,‘)

instead of (-1)"

01,

thereby

getting

x.,

(-,‘)z”

= (1

+z)--’

by the binomial theorem.) Step 3 is elementary

algebra, which yields

z(1

2;)

G(z)

-tz)(l

-z--

l+z+z2

(1 -22)(1

z)2

And that leaves us with

Ste:p

The squared factor in the denominator is a bit troublesome, since we

know that repeated roots are more complicated than distinct roots; but there

it is. We have two roots,

= 2 and

= -1; the general expansion theorem

(7.30) tells us that

~112~

+ (am +

c:l(-l)n

for some constant c, where

1+1/2+1/4 7

l-1+1

(1+1/2)2

l-2/(-1)

(The second formula for

in (7.31) is easier to use than the first one when

the denominator has nice factors. We simply substitute

= 1 /ok everywhere

R(z),

except in the factor

.where

this gives zero, and divide by (dk

)!;

this

gives the coefficient of

ndk-‘l

pk.)

Plugging in n = 0 tells us that the value of

the remaining constant c had better be

hence our answer is

= $2n+

($n+$)(-l)n.

(7.33)

It doesn’t hurt to check the cases n = 1 and 2, just to be sure that we didn’t

foul up. Maybe we should even try n = 3, since this formula looks weird. But

it’s correct, all right.

Could we have discovered (7.33) by guesswork? Perhaps after tabulating

a few more values we may have observed that g,+l

29, when n is large.

7.3 SOLVING RECURRENCES 329

And with chutzpah and luck we might even have been able to smoke out

the constant $. But it sure is simpler and more reliable to have generating

functions as a tool.

Example 3: Mutually recursive sequences.

Sometimes we have two or more recurrences that depend on each other.

Then we can form generating functions for both of them, and solve both by

a simple extension of our four-step method.

For example, let’s return to the problem of 3 x n domino tilings that we

explored earlier

this’

chapter. If we want to know only the total number of

ways, Ll,, to cover a 3 x n rectangle with dominoes, without breaking this

number down into vertical dominoes versus horizontal dominoes, we needn’t

go into as much detail as we did before. We can merely set up the recurrences

= 1 ,

Ul =o;

= 0,

=l;

=2v,-,

fl.lnp2,

LLl

vn4

)

for n 3 2.

Here

is the number of ways to cover a 3 x n rectangle-minus-corner, using

(3n

1)/2 dominoes. These recurrences are easy to discover, if we consider

the possible domino configurations at the rectangle’s left edge, as before. Here

are the values of

and V,, for small n:

nlO1234

5 6 7

(7.34)

Let’s find closed forms, in four steps. First (Step

l),

we have

= 2V,-1 +

U-2

+ [n=Ol ,

vll

b-1

+v,-2,

for all n. Hence (Step 2),

U(z) = ZzV(zj

z%l(z)+l

, V(z) =

d(z)

z2V(z)

Now (Step 3) we must solve two equations in two unknowns; but these are

easy, since the second equation yields V(z) =

zU(z)/(l

2’); we find

l-22

U(z) = --.

V(z]

l-422

+24' 1

422

(We had this formula for U(z) in (7.10), but with

instead of z2. In that

derivation, n was the number of dominoes; now it’s the width of the rectangle.)

The denominator 1

4z2 +

is a function of z2; this is what makes

I~+J

= 0 and V2, = 0, as they should be. We can take advantage of this

330 GENERATING FUNCTIONS

nice property of

by retain:ing

when we factor the denominator: We need

not take 1

4z2 +

all the way to a product of four factors (1

pkz), since

two factors of the form (1

()kz’) will be enough to tell us the coefficients. In

other words if we consider the generating function

W(z) =

l-42+z2

w()+w,z+w22+-.

we will have V(z) = zW(z’) and U(z) = (1

z2)W(z2); hence Vzn+l = W,

and

U2,,

W,,

-W,.-

We save time and energy by working with the simpler

function W(z).

The factors of 1

-4z+z1

are

(2-2-d)

and (z-2+&), and they can

also be written (1

(2+fi)z)

and (1

(2-fi)z)

because this polynomial

is its own reflection. Thus it turns out that we have

VZn+l

= wn =

3-2~6

qq2+J3)“+-(2-ti)“;

U2n =

-w,_,

3+J3

3-d

-+2+&)?-(2-\/5)n

(2+&l”

(2-m”

= 3-a

3td3

(7.37)

This is the desired closed form for the number of 3 x n domino tilings.

Incidentally, we can simplify the formula for Uzn by realizing that the

second term always lies between 0 and

The number

l-lz,,

is an integer, so

we have

(7.38)

In fact, the other term (2 --

&)n/(3

is extremely small when n is

large, because 2

0.268. This needs to be taken into account if we

try to use formula (7.38) in numerical calculations. For example, a fairly

expensive name-brand hand

Icalculator

comes up with 413403.0005 when asked

to compute (2 +

fi)‘O/(3

a).

This is correct to nine significant figures;

but the true value is slightly less than 413403, not slightly greater. Therefore

it would be a mistake to

tak.e

the ceiling of 413403.0005; the correct answer,

U20

= 413403, is obtained by rounding to the nearest integer. Ceilings can

I’ve

known slippery

be hazardous.

floors too.

Example 4: A closed form for change.

When we left the problem of making change, we had just calculated the

number of ways to pay

506.

Let’s try now to count the number of ways there

are to change a dollar, or a million dollars-still using only pennies, nickels,

dimes, quarters, and halves.

7.3 SOLVING RECURRENCES 331

The generating function derived earlier is

(qz)

1 1

F-5

pz10

pz25

-z50

this is a rational function of

with a denominator of degree 91. Therefore

we can decompose the denominator into 91 factors and come up with a 91-

term “closed form” for

C,,

the number of ways to give n cents in change.

But that’s too horrible to contemplate. Can’t we do better than the general

method suggests, in this particular case?

One ray of hope suggests itself immediately, when we notice that the

denominator is almost a function of z5. The trick we just used to simplify the

calculations by noting that 1

4z2 +

is a function of

can be applied to

C(z),

ifwe

replace

l/(1

-2)

by (1

+z-tz2+z3

+z4)/(1

-z5):

C(z)

-t

z2 +

1 1

-___--

1-S

M-5

vz10

yz25

pz50

(1+z+z2+z3+24)c(z5),

11 1 1

C(Z)

1-21-21-2~1-251-2'0'

The compressed function

c(z)

has a denominator whose degree is only 19,

so it’s much more tractable than the original. This new expression for C(z)

shows us, incidentally, that

Csn

Csn+’

C5n+2

Csn+3

C5,,+4;

and

indeed, this set of equations is obvious in retrospect: The number of ways to

leave a 53{ tip is the same as the number of ways to leave a 50# tip, because

the number of pennies is predetermined modulo 5.

Now we’re also

But

c(z)

still doesn’t have a really simple closed form based on the roots

getting compressed

reasoning.

of the denominator. The easiest way to compute its coefficients of

c(z)

probably to recognize that each of the denominator factors is a divisor of

2”.

Hence we can write

A(z)

(

) =

-zlo)5

where A(z)

=Ao+A’z+...+A3’z3’.

(7.39)

The actual value of A(z), for the curious, is

+z+...

+z~')~(1+z2+~~~+z~)(l+2~)

+2z+4z2+6z3+9z4+13z5+18z6+24z7

31z8

39z9

+ 452" +

522"

57~'~

63~'~

67~'~

69~'~

t67z"

63~'~

$57~'~

+52z20

+45z2'

39~~~

$31~~~

24~~~

t18~~~

13~~~

9z2'

6zzs

+4z29

+2z30

+z3'

332 GENERATING FUNCTIONS

Finally, since

l/(1

-z")~ = xkao

(k14)~'0k,

we can determine the coefficient

[z”]

C(z) as follows, when n = 1 Oq +

and 0 6

< 10:

lOq+r

~Aj(k:4)[10q+r=10k+jl

A:(‘:“)

+ A,+Io(‘;~) + A,+zo(~;‘) + A,+~o(‘;‘) .

(7.40)

This gives ten cases, one for each value of r; but it’s a pretty good closed

form, compared with alterrratives that involve powers of complex numbers.

For example, we can

u,se

this expression to deduce the value of

C50q

Clog.

Then

= 0 and we have

c50q =

("k")

+45(q;3)+52(4;2)

+2(“3

The number of ways to change 50# is (i)

+45(t)

= 50; the number of ways

to change $1 is

($)

+45(i)

-t 52(i) = 292;

and the number of ways to change

$l,OOO,OOO

= 66666793333412666685000001.

Example 5: A divergent series.

Now let’s try to get a closed form for the numbers

defined by

= 1;

n =

ngv1,

for

> 0.

After staring at this for a Sew nanoseconds we realize that

g,,

is just n!; in

Nowadayspeo-

fact, the method of summation factors described in Chapter 2 suggests this

~~~~‘e~c~~~

answer immediately. But let’s try to solve the recurrence with generating

functions, just to see what happens. (A powerful technique should be able to

handle easy recurrences like this, as well as others that have answers we can’t

guess so easily.)

The equation

ngn-1

[n=Ol

holds for all n, and it leads to

G(z) =

xgnz”

~ng,-rz”+~z’.

n n

n=O

To complete Step 2, we want to express

ng,

1 2” in terms of G(z), and the

basic maneuvers in Table 320 suggest that the derivative G’(z) =

ngnzn

’

7.3 SOLVING RECURRENCES 333

is somehow involved. So we steer toward that kind of sum:

G(z)

l+t(n+l)g,M+’

ng, zn+l +

zn+’

= 1

+z’G’(z)+zG(z).

Let’s check this equation, using the values of

g,,

for small n. Since

G = 1

+z+2z2

6z3

+24z4

...

G’

1+42

+18z2+96z3+-.,

we have

z2G’

z2+4z3+18z4+96z5+.-,

zG =

z+z2

+2z3

6z4

+24z5

...

1 = 1.

These three lines add up to G, so we’re fine so far. Incidentally, we often find

it convenient to write ‘G’ instead of ‘G(z)‘; the extra

‘(2)’

just clutters up the

formula when we aren’t changing

Step 3 is next, and it’s different from what we’ve done before because we

have a differential equation to solve. But this is a differential equation that

we can handle with the hypergeometric series techniques of Section 5.6; those

techniques aren’t too bad. (Readers who are unfamiliar with hypergeometrics

“This will be ouick.”

needn’t worrv- this will be quick.)

That’s what

the

doctor said just

First we must

before he

stuck

both sides:

get rid of the constant

‘l’,

so we take the derivative of

with that needle.

Come to think of it,

“hypergeometric”

sounds a lot like

G’ = @‘G’S zG

1)’ =

(2zG’+z’G”)+(G

+zG’)

z2G”+3zG’+G.

“hypodermic.”

The theory in Chapter 5 tells us to rewrite this using the 4 operator, and we

know from exercise 6.13 that

= zG’,

B2G

z2G”

+zG’.

Therefore the desired form of the differential equation is

4G =

~9~G+224G+zG

z(9+1)‘G.

According to

(5.1og),

the solution with

= 1 is the hypergeometric series

F(l,l;;z).

334 GENERATING FUNCTIONS

Step 3 was more than we bargained for; but now that we know what the

function G is, Step 4 is easy-the hypergeometric definition (5.76) gives us

the power series expansion:

We’ve confirmed the closed

:form

we knew all along,

g,,

= n!.

Notice that the technique gave the right answer even though G(z) di-

verges for all nonzero z. The sequence n! grows so fast, the terms In!

zTll

approach

0;)

as n -+ 00,

un:less

= 0. This shows that formal power series

can be manipulated algebraically without worrying about convergence.

Example 6: A recurrence that goes ail the way back.

Let’s close this section by applying generating functions to a problem in

graph theory. A fun of order n is a graph on the vertices {0,

. . . , n} with

1 edges defined as follows: Vertex 0 is connected by an edge to each of

the other n vertices, and vertex k is connected by an edge to vertex k +

for

1 6 k < n. Here, for example, is the fan of order 4, which has five vertices

and seven edges.

The problem of interest: How many spanning trees

are in such a graph?

A spanning tree is a subgraph containing all the vertices, and containing

enough edges to make the subgraph connected yet not so many that it has

a cycle. It turns out that every spanning tree of a graph on n + 1 vertices

has exactly n edges. With fewer than n edges the subgraph wouldn’t be

connected, and with more

t:han

n it would have a cycle; graph theory books

prove this.

There are (‘“L’) ways to choose n edges from among the 2n

1 present

in a fan of order n, but these choices don’t always yield a spanning tree. For

instance the subgraph

has four edges but is not a spanning tree; it has a cycle from 0 to 4 to 3 to 0,

and it has no connection between

,2} and the other vertices. We want to

count how many of the (‘“i ‘) choices actually do yield spanning trees.

7.3 SOLVING RECURRENCES 335

Let’s look at some small cases. It’s pretty easy to enumerate the spanning

trees for n =

2, and 3:

&I!

f, = 1

= 3

= 8

(We need not show the labels on the vertices, if we always draw vertex 0 at

the left.) What about the case n = O? At first it seems reasonable to set

fc = 1; but we’ll take

= 0, because the existence of a fan of order 0 (which

should have 2n

1 = -1 edges) is dubious.

Our four-step procedure tells us to find a recurrence for

that holds for

all n. We can get a recurrence by observing how the topmost vertex (vertex n)

is connected to the rest of the spanning tree. If it’s not connected to vertex 0,

it must be connected to vertex n

since it must be connected to the rest of

the graph. In this case, any of the

f,-

1 spanning trees for the remaining fan

(on the vertices 0 through n

1) will complete a spanning tree for the whole

graph. Otherwise vertex n is connected to 0, and there’s some number k < n

such that vertices n, n-

. . , k are connected directly but the edge between

k and k

1 is not in the subtree. Then there can’t be any edges between

0 and {n

1,. . . , k}, or there would be a cycle. If k =

the spanning tree

is therefore determined completely. And if k > 1, any of the fk-r ways to

produce a spanning tree on {0,

. . .

, k

l} will yield a spanning tree on the

whole graph. For example, here’s what this analysis produces

when n = 4:

k=4 k=3 k=2

k=l

f4 f3 f3

The general equation, valid for n 2

fn = f,-1

f,-1 +

f,-1

+ fn-3

. . + f, + 1 .

(It almost seems as though the ‘1’ on the end is

and we should have chosen

= 1; but we will doggedly stick with our choice.) A few changes suffice to

make the equation valid for all integers n:

f,-j

+ [n>O] .

kin

336 GENERATING FUNCTIONS

This is a recurrence that “goes all the way back” from

f,-l

through all pre-

vious values, so it’s different from the other recurrences we’ve seen so far

in this chapter. We used a special method to get rid of a similar right-side

sum in Chapter 2, when we solved the quicksort recurrence (2.12); namely,

we subtracted one instance of the recurrence from another (f,+l

fn).

This

trick would get rid of the

now, as it did then; but we’ll see that generating

functions allow us to work directly with such sums. (And it’s a good thing

that they do, because we will be seeing much more complicated recurrences

before long.)

Step 1 is finished; Step

is where we need to do a new thing:

F(z)

tf,zn

tf,,zn+tfkzn[k<n]+t(n>O]zn

n n

zF(z)

~fkZk~[n>k]Znpk

k n

= zF(z) + F(z)

m>O

zF(z)

F(z)

it-.

1-z

The key trick here was to change zn to z

n-k;

this made it possible to express

the value of the double sum in terms of F(z), as required in Step 2.

Now Step 3 is simple algebra, and we find

F(z)

’

-3zf22

Those of us with a zest for memorization will recognize this as the generating

function (7.24) for the even-numbered Fibonacci numbers. So, we needn’t go

through Step 4; we have found a somewhat surprising answer to the

spans-

of-fans problem:

F2n

for n 3 0.

(

7.42)

7.4

SPECIAL GENERATING FUNCTIONS

Step 4 of the four-step procedure becomes much easier if we know

the coefficients of lots of

diff’erent

power series. The expansions in Table 321

are quite useful, as far as they go, but many other types of closed forms are

possible. Therefore we ought to supplement that table with another one,

which lists power series that correspond to the “special numbers” considered

in Chapter 6.

7.4 SPECIAL GENERATING FUNCTIONS 337

Table 337 Generating functions for special numbers.

xWm+n-Hml

(7.43)

z)m+l

l-z

Fmz

(Frn-l+Fm+l)~+ (-l)mz2

(p)"

(1-z)(l-222;1...(1-mz)

iii

= z(z+ 1). .

.(z+m-1)

2”

(e’-

1)” =

rn!&tz}$

z+wz

e _

wm-

m.n>O

ewieL-l)

wm-

m,n>O

(1=

wm-

m,nbO

l-w

e(W-l)z

wm-

m,n>O

(7.44)

(7.45)

(7.46)

(7.47)

(7.48)

(7.49)

(7.50)

(7.51)

(7.52)

(7.53)

(7.54)

(7.55)

(7.56)

338 GENERATING FUNCTIONS

Table 337 is the database we need. The identities in this table are not

difficult to prove, so we needn’t dwell on them; this table is primarily for

reference when we meet a new problem. But there’s a nice proof of the first

formula, (7.43), that deserves mention: We start with the identity

(xy)zn

-2)x+’

and differentiate it with respect to x. On the left, (1

z)-~-’

is equal to

elx+l~ln~llll-rll

so d/dx contributes a factor of

ln(l/(

2)). On the right,

the numerator

(“‘-,“)

is (x +n) . . . (x + 1

and d/dx splits this into n terms

whose sum is equivalent to :multiplying

(“‘,“)

1 1

-+...+-

x+n

x+1

x+Tl

H, .

Replacing x by m gives (7.43). Notice that H,+n

H, is meaningful even

when x is not an integer.

By the way, this method of differentiating a complicated product

leav-

ing it as a product-is usually better than expressing the derivative as a sum.

For example the right side of

$(i

x+n)“...(x+l)‘)

(x+n)n...(x+l)’

(

*+...+A

would be a lot messier written out as a sum.

The general identities in Table 337 include many important special cases.

For example, (7.43) simplifies to the generating function for

when m = 0:

&ln&

tH,z”.

(7.57)

This equation can also be derived in other ways; for example, we can take the

power series for

ln(l/(

z))

and divide it by 1

z to get cumulative sums.

Identities (7.51) and (7.52) involve the respective ratios

{,~,}/(“~‘)

and

[,“‘J

/(“c’),

which have the undefined form O/O when n 3 m. However,

there is a way to give them a proper meaning using the Stirling polynomials

of (6.45), because we have

{mmn}/(m~l)

(-l)n+‘n!mo,(n-m);

[m~n]/(m~l)

n!mo,(m).

(7.59)

7.4 SPECIAL GENERATING FUNCTIONS 339

Thus, for example, the case

= 1 of (7.51) should not be regarded as the

power series

,&,O(zn/n!){,

l,}/(z),

but rather as

ln(1 +

-t(-z)“oll(n-l)

= 1

+~z-~zz+...

II20

Identities

(7.53), (7.551, (7.54),

and (7.56) are “double generating func-

tions” or “super generating functions” because they have the form G (w,

t,,,

Sm,n

~“‘2~.

The coefficient of wm is a generating function in the vari-

able

the coefficient of 2” is a generating function in the variable w.

7.5 CONVOLUTIONS

always thought

convolution was

what happens to

my brain when

try to do a proof.

The convolution of two given sequences

(fo,

, . . ) =

(f,,)

and

(SOlSl,. .

(gn)

the

sequence

(f0g0,

fog1 +

flg0,

. .

(xkfkgn

k).

We have observed in Sections 5.4 and 7.2 that convolution of sequences cor-

responds to multiplication of their generating functions. This fact makes it

easy to evaluate many sums that would otherwise be difficult to handle.

Example 1: A Fibonacci convolution.

For example, let’s try to evaluate

~~=,

FkFn~-k

in closed form. This is

the convolution of (F,) with itself, so the sum must be the coefficient of 2”

F(z)',

where

F(z)

is the generating function for (F,). All we have to do is

figure out the value of this coefficient.

The generating function

F(z)

z/(

-z-z’),

a quotient of polynomials; so

the general expansion theorem for rational functions tells us that the answer

can be obtained from a partial fraction representation. We can use the general

expansion theorem (7.30) and grind away; or we can use the fact that

Instead of expressing the answer in terms of

and

$i,

let’s try for a closed

form in terms of Fibonacci numbers. Recalling that

+ $ =

we have

$“+$” =

[z”l

(

(Q+$)z

Lz"'

($z)(

qjz)

2-z

l-Z-22

2F,+,

-F,.

340 GENERATING FUNCTIONS

F(z)'

(n + 1 )(;!F,+r

-F,,)2-

;

F,+I

.zn

7x30

Tl30

and we have the answer we seek:

FkFn-k

2nF,+ 1

--(n+l)F,

k=O

(7.60)

For example, when n = 3 this formula gives

F,JF~

FlF2

FzFl

F~F,J

0+ 1 +1 +0

on the left and

(6F4

-4F3)/5

q = (18-8)/5

on the right.

Example 2: Harmonic convolutions.

The efficiency of a certain computer method called “samplesort” depends

on the value of the sum

integers

m,n

3 0.

Exercise 5.58 obtains the value of this sum by a somewhat intricate double

induction, using summation factors. It’s much easier to realize that

Tm,n

just the nth term in the convolution of

((i),

(A),

(i),

. . .) with (0, $,

. .

.).

Both sequences have simple generating functions in Table 321:

zn =

--.

-z)nl+l

ln&.

n>O

Therefore, by (7.43),

1 1

m,n

[z”l

_“,,,,l

1-z

'z"-"'

-Z)m+l

U-k’-LJ

nnm

(

In fact, there are many more sums that boil down to this same sort of

convolution, because we have

-z)'+s+2

In-

for all

and s. Equating coefficients of 2” gives the general identity

;

(‘:“)

(s+nl;k)IH,+d-&)

(r+s+n,L+l)(H.+s+n+~

-H,+,+I)

(7.61)

7.5 CONVOLUTIONS 341

Beta

use it’s so

harmonic.

This seems almost too good to be true. But it checks, at least when n = 2:

(T+;+3)(r+:+3+r+j+2)

Special cases like

0 are as remarkable as the general case.

And there’s more. We can use the convolution identity

(‘:“)(“fn”*k)

(r+y+‘)

to transpose H, to

t,he

other side, since H, is independent of k:

;

(r;k)(s;:;k)Hr+~

(I+sfn+‘)(Hr+rni~

-H,+,+,

+H,).

There’s still more: If

and

are nonnegative integers

and m, we can replace

(‘+kk)

by (‘I”) and (“‘,“i”) by (‘“‘,“Pk); then we can change k to k-

and

n to n

gett,ing

integers

m, n 3 0.

(7.63)

Even the special case

m = 0 of this identity was difficult for us to handle

in Chapter

(See (2.36).) We’ve come a long way.

Example 3: Convolutions of convolutions.

If we form the convolution of (fn) and (g,,), then convolve this with a

third sequence (h,), we get a sequence whose nth term is

The generating function of this three-fold convolution is, of course, the three-

fold product F(z) G(z) H(z). In a similar way, the m-fold convolution of a

j+k+l=n

sequence (

gn)

with itself has nth term equal to

gk, gkl

...

gk,

+kr+...+k,=n

and its generating function is

Go.

342 GENERATING FUNCTIONS

We can apply these observations to the spans-of-fans problem considered

earlier (Example 6 in Section 7.3). It turns out that there’s another way to

compute

f,,

the number of spanning trees of an n-fan, based on the config-

urations of tree edges between the vertices

{1,2,.

. . , n}: The edge between

vertex k and vertex k + 1 may or may not be selected for the subtree; and

each of the ways to select these edges connects up certain blocks of adjacent

Concrete blocks.

vertices. For example, when n = 10 we might connect vertices

{1,2},

{3},

{4,5,6,7}, and {8,9,10}:

How many spanning trees can we make, by adding additional edges to ver-

tex O? We need to connect 0 to each of the four blocks; and there are two

ways to join 0 with

{1,2},

one way to join it with

{3},

four ways with {4,5,6,7},

and three ways with {S, 9,

lo},

or 2 9 1

.4.3

= 24 ways altogether. Summing

over all possible ways to make blocks gives us the following expression for the

total number of spanning trees:

fn=E

k,kz...k,.

(7.64)

m>O

+kz+...+k,=n

,kJ,...,k,>O

Forexample,

=4+3~1+2~2+1~3+2~1~1+1~2~1+1~1~2+1~1~1~1

=21.

This is the sum of m-fold convolutions of the sequence (0,

1,2,3,.

. .

for

m=l,

2,3,

. . . . hence the generating function for (fn) is

F(z)

= G(z)+

G(z)'+

+...

,';',21,)

where G(z) is the generating function for (0,

1,2,3,.

.),

namely

z/(1

2)'.

Consequently we have

F(z) =

(,_;2+

l-32+22'

as before. This approach to

(f,,)

is more symmetrical and appealing than the

complicated recurrence we had earlier.

7.5 CONVOLUTIONS 343

Example 4: A convoluted recurrence.

Our next example is especially important; in fact, it’s the “classic exam-

ple” of why generating functions are useful in the solution of recurrences.

Suppose we have n + 1 variables

x0,

x1,

. . . , x, whose product is to be

computed by doing n multiplications. How many ways

are there to insert

parentheses into the product

‘x1

. . .

:x,

so that the order of multiplication is

completely specified? For example, when n = 2 there are two ways, xc.

(xl

.x2

)

and

(x0.x,

) .

x2.

And when n = 3 there are five ways,

Thus

= 2,

= 5; we also have

= 1 and

= 1.

Let’s use the four-step procedure of Section 7.3. What is a recurrence

for the C’s? The key observation is that there’s exactly one

‘

’

operation

outside all of the parentheses, when n > 0; this is the final multiplication

that ties everything together. If this

’

occurs between

and

xk+l

, there

are

ways to

full,y

parenthesize xc.. . . .

Xk,

and there are C,-

1 ways to

fully parenthesize

Xk+l

. . . . x,; hence

CoC,-l+C,C,~2+~~'+C,~,C~,

ifn>O.

By now we recognize this expression as a convolution, and we know how to

patch the formula so that it holds for all integers n:

xCkCn-l-k

[n=o].

(7.65)

Step 1 is now complete. Step 2 tells us to multiply by

Z”

and sum:

C(z)

c,zn

ckcn-,

-kZn

k.n

n=O

ckZkx

cn-,-kZn-k

c(z)~zc(z)+

The authors jest.

Lo and behold, the convolution has become a product, in the generating-

function world. Life is full of surprises.

344 GENERATING FUNCTIONS

Step 3 is also easy. We solve for

C(z)

by the quadratic formula:

C(z)

1*di-=G

But should we choose the + isign or the

sign? Both choices yield a function

that satisfies C(z) =

K(z)’

-1-

but only one of the choices is suitable for our

problem. We might choose the + sign on the grounds that positive thinking

is best; but we soon discover that this choice gives C(0) = 00, contrary to

the facts. (The correct function C(z) is supposed to have C(0) =

= 1.)

Therefore we conclude that

1-Jl-42

C(z)

Finally, Step 4. What is

[zn]

C(z)? The binomial theorem tells us that

k>O

(‘f)

(-4zjk

= 1 +

(rl/Y)

(-4z)k

;

hence, using (5.37),

(--‘/‘>~

(;)A$

nao

ll)O

The number of ways to parenthesize, C,, is (‘,“)

We anticipated this result in Chapter 5, when we introduced the sequence So

the

convo-

of Catalannumbers

(1,1,2,5,14,.

. . ) = (C,). This sequence arises in dozens luted

recurrence

of problems that seem at first to be unrelated to each other

[41],

because

has led us to an

many situations have a recursive structure that corresponds to the convolution

oft-recurring

con-

volution.

recurrence (7.65).

For example, let’s consider the following problem: How many sequences

(al,a2..

. , al,,) of

+1's

and

-1's

have the property that

+ a2

. . +

azn =

and have all their partial sums

al,

+a2,

. . . .

+a2+...+aZn

nonnegative? There must be n occurrences of

and n occurrences of -1.

We can represent this problem graphically by plotting the sequence of partial

7.5 CONVOLUTIONS 345

sums s, = XL=,

as a function of n: The five solutions for n = 3 are

These are “mounta.in ranges” of width

that can be drawn with line seg-

ments of the forms /and

It turns out that there are exactly

ways to

do this, and the sequences can be related to the parenthesis problem in the

following way: Put an extra pair of parentheses around the entire formula, so

that there are n pairs of parentheses corresponding to the n multiplications.

Now replace each

‘

’

by +1 and each

‘

)

’

by -1 and erase everything else.

For example, the formula x0.

((xl

.x1).

(xs .x4)) corresponds to the sequence

(+l,+l,-l,+l,+l,-1,-1,-l) by this rule. The five ways to parenthesize

x0 .x1

.x2.

x3 correspond to the five mountain ranges for n = 3 shown above.

Moreover, a slight reformulation of our sequence-counting problem leads

to a surprisingly simple combinatorial solution that avoids the use of gener-

ating functions: How many sequences

(ao,

al,

al,.

. . , azn) of

+1's

and -1's

have the property that

a0 + al + a2 + . . . + azn = 1 ,

when all the partial sums

a0,

al,

a0+al

+a2,

. . . .

a0 +

+ . . + azn

are required to be positive? Clearly these are just the sequences of the pre-

vious problem, with the additional element

placed in front. But

the sequences in

th.e

new problem can be enumerated by a simple counting

argument, using a remarkable fact discovered by George Raney

[243]

in 1959:

If(x,,xz,...

, x,) is any sequence of integers whose sum is

, exactly one of

the cyclic shifts

(x1,x2,...

,xrn),

(XZ!...,&n,Xl),

'..,

(Xtn,Xl,...,&l-1)

has all of its partial sums positive.

For example, consider the sequence

(3,

-5,2,

-2,3,0).

Its cyclic shifts are

(3,

-5,2,

-2,310)

(-A&O,&-5,4

(-5,2,

-2,3,0,3)

(3,0,3,

-5,2,

-2) J

(2,

-2,3,0,3,

-5) (0,3,

-5,2,

-2,3)

and only the one that’s checked has entirely positive partial sums.

346 GENERATING FUNCTIONS

Raney’s lemma can be proved by a simple geometric argument. Let’s

extend the sequence periodically to get an infinite sequence

thus we let X,+k = xk

for

a.11 k 3 0. If we now plot the partial sums s, =

...

as a function

Iof

n, the graph of s, has an “average slope” of

l/m, because

s,+,,

s,,

For example, the graph corresponding to our

example sequence (3,

-5,2,

--2,3,0,3,

-5,2,.

. . ) begins as follows:

The entire graph can be comained between two lines of slope 1

/m,

as shown;

we have m = 6 in the illustration. In general these bounding lines touch the

graph just once in each cycle of m points, since lines of slope l/m hit points

with integer coordinates only once per m units. The unique lower point of

intersection is the only place in the cycle from which all partial sums will

be positive, because every other point on the curve has an intersection point

within m units to its right.

With Raney’s lemma we can easily enumerate the sequences

(ao,

. . . , aln)

of +1’s and -1’s whose partial sums are entirely positive and whose total

sum is

There are (‘“,“)

sequences with n occurrences of -1 and n + 1

occurrences of

+l,

and Raney’s lemma tells us that exactly

1/(2n

+ 1) of

these sequences have all partial sums positive. (List all N = (‘“,“) of these

sequences and all 2n + 1 of

t:heir

cyclic shifts, in an N x (2n + 1) array. Each

row contains exactly one solution. Each solution appears exactly once in each

column. So there are

N/(2ni-1)

distinct solutions in the array, each appearing

(2n + 1) times.) The total number of sequences with positive partial sums is

Ah, if stock prices

would only continue

to rise like this.

(Attention, com-

puter scientists:

The partial sums

in this problem

represent the stack

size as a function of

time, when a prod-

uct of n + 1 factors

is evaluated, be-

cause each “push”

operation changes

the size by +1 and

each multiplication

changes it by -1 .)

Example 5: A recurrence with m-fold convolution.

We can generalize the problem just considered by looking at sequences

(a0,. . .

amn)

of +1’s and (1

m)‘s whose partial sums are all positive and

7.5 CONVOLUTIONS 347

(Attention, com-

puter scientists:

The stack interpre-

tation now applies

with respect to an

m-ary operation,

instead of the bi-

nary multiplication

considered earlier.)

whose total sum is

. Such sequences can be called m-Raney sequences. If

there are k occurrences of (1

m) and mn + 1

k occurrences of

, we have

k(l-m)+(mn+l-k) = 1,

hence k = n. There are

(“t+‘)

sequences with n occurrences of (1

m) and

mn + 1

n occurrences of

+l,

and Raney’s lemma tells us that the number

of such sequences with all partial sums positive is exactly

mn+l

(

-- =

n mn+l n

(m-l)n+l’

(7.66)

So this is the number of m-Raney sequences. Let’s call this a Fuss-Catalan

number

Cim,“‘,

because the sequence

(&“‘)

was first investigated by N.I.

Fuss

[log]

in 1791 (many years before Catalan himself got into the act). The

ordinary Catalan numbers are

Cr’.

Now that we

k:now

the answer, (7.66), let’s play “Jeopardy” and figure

out a question that leads to it. In the case m = 2 the question was: “What

numbers

satisfy the recurrence

CkCnPiPk + (n = O]?” We will

try to find a similar question (a similar recurrence) in the general case.

The trivial sequence (+1) of length 1 is clearly an m-Raney sequence. If

we put the number (1 -m) at the right of any m sequences that are m-Raney,

we get an m-Raney sequence; the partial sums stay positive as they increase

to +2, then +3, . . . ,

fm,

and

. Conversely, we can show that all m-Raney

sequences (ae, . . . , ~a,,) arise in this way, if n > 0: The last term a,,,,, must

be (1

m). The partial sums

. +

aj-

1 are positive for 1 < j 6 mn,

and

s,,

= m because

s,,

+ a,,,,, = 1. Let kl be the largest index 6 mn such

that

Sk,

= 1; let

be largest such that

skz

= 2; and so on. Thus

ski

= j

and

> j, for ki

k 6 mn and 1 < j 6 m. It follows that

= mn, and

we can verify without difficulty that each of the subsequences (ae, . . . , ok,

-I),

(ok,, . . . , okJPi), . . . , (ok,,-, , . . . ,

ok,,,

-1) is an m-Raney sequence. We must

have kl = mnl + 1,

kl = mn2 + 1, . . . ,

k,_l

= mn, +

for

some nonnegative integers

nl,

n2, . . . , n,.

Therefore (“‘t’-‘)

is the answer to the following two interesting ques-

tions: “What are the numbers Cim’ defined by the recurrence

for all integers n?”

“If G(z) is a power series that satisfies

G(z)

zG(z)"

+ 1, (7.68)

what is

[z”]

G(z)?”

348 GENERATING FUNCTIONS

Notice that these are not easy questions. In the ordinary Catalan case

(m = 2), we solved (7.68) for G(z) and its coefficients by using the quadratic

formula and the binomial theorem; but when m = 3, none of the standard

techniques gives any clue about how to solve the cubic equation G = zG3 + 1.

So it has turned out to be easier to answer this question before asking it.

Now, however, we know enough to ask even harder questions and deduce

their answers. How about this one: “What is

[z”]

G(z)‘, if

is a positive

integer and if G(z) is the power series defined by (7.68)?” The argument we

just gave can be used to show that [PI G(z)’ is the number of sequences of

length mn +

with the following three properties:

Each element is either

$-1

or (1

m).

The partial sums are all positive.

The total sum is

For we get all such sequences in a unique way by putting together

sequences

that have the m-Raney property. The number of ways to do this is

c’m’c’m’

t..

CL:)

[znl

G(z)'.

+nr

t...+n,=n

Raney proved a generalization of his lemma that tells us how to count

such sequences: If

(XI,

x2,. . . ,

x,)

is any sequence of integers with

for

all j, and with

+ x2 + . . .

-1

> 0,

exactly

of the cyclic shifts

(x1,x2,..

.,xm), (X2,...,Xm,Xl), .

..1

(%il,Xl,... ,xTn 1

)

have all positive partial sums.

For example, we can check this statement on the sequence

(-2,1,

-l,O,

l,l,-l,l,l,l).

The cyclic shifts are

(-2,1,-l,O,l,l,-l,l,l,l)

(1,~l,l,l,l,-2,1,-l,O,l)

(l,-l,O,l,l,-l,l,l,l,--2)

(-l,l,l,l,-2,1,-l,O,l,l)

(-l,O,l,l,-l,l,l,l,-2,l)

(l,l,l,-2,1,-1,0,1,1,-l)

(O,l,l,-1,1,1,1,-2,1,--l)

(l,l,-2,1,-l,O,l,l,-1,l)

(1,1,-~,1,1,1,-2,1,~1,0)

(l,-2,1,-l,O,l,l,-l,l,l)

and only the two examples marked

‘J’

have all partial sums positive. This

generalized lemma is proved in exercise 13.

A sequence of

+1's

and (1

m)‘s that has length mn+

and total sum

must have exactly n occurrences of (1

m). The generalized lemma tells

us that

L/(mn

of these

‘“‘t+‘)

sequences have all partial sums positive;

7.5 CONVOLUTIONS 349

hence our tough question has a surprisingly simple answer:

[znl

G(z)’

(“I+‘)

for all integers

> 0.

Readers who haven’t forgotten Chapter 5 might well be experiencing

dkjjh

vu:

“That formula looks familiar; haven’t we seen it before?” Yes, indeed;

equation (5.60) says that

[z”]B,(z)’

-Jr

(

)

Therefore the generating function G(z) in (7.68) must actually be the gener-

alized binomial series

‘B,(z).

Sure enough, equation (5.59) says

cBm(z)‘-m

Tim(z)-”

= 2)

which is the same as

T3B(z)-l = zB,(z)"

Let’s switch to the notation of Chapter 5, now that we know we’re dealing

with generalized binomials. Chapter 5 stated a bunch of identities without

proof. We have now closed part of the gap by proving that the power series

IBt

(z) defined by

TQ(z)

(

has the remarkable property that

%(z)’

(yr)$&,

whenever t and

;Ire

positive integers.

Can we extend these results to arbitrary values oft and

I-?

Yes; because

the coefficients (t:T’)

are polynomials in t and

The general rth power

defined by

‘B,(z)’ = e

rln’Bt(z)

-9

rln93t(z))n

ll20

(I-y)nl)‘,

ll>O

llI>l

has coefficients that are polynomials in t and r; and those polynomials are

equal to

(tnn+‘)

for infinitely many values oft and r. So the two sequences

of polynomials must be identically equal.

350 GENERATING FUNCTIONS

Chapter 5 also mentions the generalized exponential series

which is said in (5.60) to hzve an equally remarkable property:

[z”]

Et(=)’

etn

+-,w

We can prove this as a limiting case of the formulas for

‘BBt

(z), because it is

not difficult to show that

7.6

EXPONEN’I’IAL GF’S

Sometimes a sequence

(gn)

has a generating function whose proper-

ties are quite complicated, while the related sequence

(g,/n!)

has a generating

function that’s quite simple. In such cases we naturally prefer to work with

(gJn!)

and then multiply by n! at the end. This trick works sufficiently

often that we have a special name for it: We call the power series

(7.71)

the exponential generating function or

‘<egfr’

of the sequence

(go,

gl,

g2,

. . .

This name arises because the exponential function

is the egf of (1 , 1 , 1, . , .

Many of the generating functions in Table 337 are actually egf’s. For

example, equation (7.50) says that (In

&)m/m!

is the egf for the sequence

([:I,

[:I,

[:]d.

ordinary generating function for this sequence is

much more complicated (and also divergent).

Exponential generating

:functions

have their own basic maneuvers, analo-

gous to the operations we learned in Section 7.2. For example, if we multiply

the egf of

(gn)

by z, we get

&.n+l

Sn-

iYE

w-1

;

n>O

n>l

G-l

j&y

n>O

this is the egf of (0, go,Zgl, . . .) =

(ng,-1).

Differentiating the egf of (go, 91,

g2,

. . . ) with respect to z gives

Are we having

fun yet?

(7.72)

7.6 EXPONENTIAL GENERATING FUNCTIONS 351

this is the egf of

(g-1,

g2,. . .

Thus differentiation on egf’s corresponds to the

left-shift operation (G(z) ~ go)/z on ordinary gf’s. (We used this left-shift

property of egf’s when we studied hypergeometric series, (5.106).) Integration

of an egf gives

g,,;dt

(7.73)

this is a right shift, the egf of (0,

go,

91). .

.).

The most interesting operation on egf’s, as on ordinary gf’s, is multipli-

cation. If

i(z)

and G(z) are egf’s for (f,,) and (gn), then

i(z)G(z)

= A(z) is

the egf for a sequence (hn) called the binomial convolution of (f,,) and (g,,):

Binomial coefficients appear here because (z) = n!/k! (n ~ k)!, hence

in other words,

(h,/n!)

is the ordinary convolution of

(f,,/n!)

and (g,,/n!).

Binomial convolutions occur frequently in applications. For example, we

defined the Bernoulli numbers in (6.79) by the implicit recurrence

[m=O],

for all m 3 0;

this can be rewritten as a binomial convolution, if we substitute n for m + 1

and add the term

ES,

to both sides:

B,+[n=l],

for all n 3 0.

We can now relate this recurrence to power series (as promised in Chapter 6)

by introducing the egf for Bernoulli numbers, B(z) = EnSo B,,z’/n!. The

left-hand side of (7.75) is the binomial convolution of (B,,) with the constant

sequence (1 , 1 ,

);

hence the egf of the left-hand side is

z)e’. The egf

of the right-hand side is

Ena

(B, +

[n=l])z”/n!

= B(z) + z. Therefore we

must have B(z) = z/(e’ ~ 1); we have proved equation (6.81), which appears

also in Table 337

a:s

equation (7.44).

352 GENERATING FUNCTIONS

Now let’s look again at a sum that has been popping up frequently in

this book,

S,(n) =

+ 1

+ 2”’

. . + (n

1)” =

km.

O<k<n

This time we will try to analyze the problem with generating functions, in

hopes that it will suddenly become simpler. We will consider n to be fixed

and m variable; thus our goal is to understand the coefficients of the power

series

S(z)

= S0(n)+Sl(n)z+S2(n)z2+~~~ =

Sm(n)zm.

ma0

We know that the generating function for (1, k, k2, . . . ) is

1 -kz

kmzm,

m>O

hence

S(z) =

kmzfn

ma0 O<k<n O<k<n

’

by interchanging the order of summation. We can put this sum in closed

form,

(7.76)

but we know nothing about expanding such a closed form in powers of z.

Exponential generating functions come to the rescue. The egf of our

sequence (Sc(n),Sr(n),Sz(n),...) is

S(z,n) = So(n)

+Sl(n)

+Sz(n)

f...

S,(n)

m30

To get these coefficients

S,(n)

we can use the egf for (1, k, k2,. . .

namely

$2 =

ma0

km$,

and we have

S(z,n) =

ekz.

m>O O$k<n O$k<n

7.6 EXPONENTIAL GENERATING FUNCTIONS 353

And the latter

sumI

is a geometric progression, so there’s a closed form

S(z,n) =

$+.

(7.77)

All we need to do is figure out the coefficients of this relatively simple function,

and we’ll know

S,i:n),

because S,(n) = m! [z”‘]S(z,n).

Here’s where 13ernoulli numbers come into the picture. We observed a

moment ago that

t.he

egf for Bernoulli numbers is

hence we can write

enz-1

S(z) = B(z) --

Bo~.+B,~+Bz~+...)(n~+n2~+n3~+-..)

(

The sum S,(n) is m! times the coefficient of

z”’

in this product. For example,

So(n)

(h3&)

S(n)

(

Elom+Blm

= .!n2-d-n;

f%(n)

.2!

Bo$

(

in3-tn2+in.

. . . . * .

We have therefore derived the formula

= Sz(n) =

$n(n

i)(n

1) for

the umpteenth time, and this was the simplest derivation of all: In a few lines

we have found the general behavior of S,(n) for all m.

The general fo:rmula can be written

%-l(n)

&EL,(n)

B,(O))

where B,(x) is the Bernoulli polynomial defined by

(7.78)

B,(x) =

(;)BkX-‘.

(7.79)

Here’s why: The Bernoulli polynomial is the binomial convolution of the

sequence

(Bo,

B1,

B;r,

. . . ) with (1,

x,x2,.

. . ); hence the exponential generating

354 GENERATING FUNCTIONS

function for (Be(x),

(x),

(x), . . .) is the product of their egf’s,

@2,x)

B,,,(x)2

-?.-

zexz

ez-

m>O

=I-.

(7.80)

In>0

Equation (7.78) follows because the egf for (0, So(n),

(n), . . .

)

is, by (7.77),

z- = B(z,n)

B(z,O)

Let’s turn now to another problem for which egf’s are just the thing:

How many spanning trees are possible in the complete graph on n vertices

{1,2,...

, n}? Let’s call this number t,,. The complete graph has

$(n

edges, one edge joining

each

pair of distinct vertices; so we’re essentially

looking for the total number of ways to connect up n given things by drawing

1 lines between them.

We have tl =

= 1. Also

= 3, because a complete graph on three

vertices is a fan of order 2; we know that

= 3. And there are sixteen

spanning trees when n = 4:

I/IL-I

Ia:

(7.81)

Hence

= 16.

Our experience with the analogous problem for fans suggests that the best

way to tackle this problem is to single out one vertex, and to look at the blocks

or components that the spanning tree joins together when we ignore all edges

that touch the special vertex. If the non-special vertices form

components

of sizes kl , kz,

, . . ,

k,,

then we can connect them to the special vertex in

klk2..

ways. For example, in the case n = 4, we can consider the lower

left vertex to be special. The top row of (7.81) shows

3t3

cases where the other

three vertices are joined among themselves in

ways and then connected to

the lower left in 3 ways. The bottom row shows 2.1 x tztl x (i) solutions where

the other three vertices are divided into components of sizes 2 and 1 in (i)

ways; there’s also the case

where the other three vertices are completely

unconnected among themselves.

This line of reasoning leads to the recurrence

n-l

kl,kz,...,k,

klk2...k,tk,tkz

..tk.

’

k,+kz+...+k,=n-1

7.6 EXPONENTIAL GENERATING FUNCTIONS 355

for all n > 1.

Here”s

why: There are

(k,

,:,,‘,,k_) ways to assign n- 1 elements

to a sequence of

TTL

components of respective sizes

kl,

k2, . . . , k,; there are

tk, tk1 . . .

tk,

ways to connect up those individual components with spanning

trees; there are

k.2

. . .

ways to connect vertex n to those components; and

we divide by m! because we want to disregard the order of the components.

For example, when n = 4 the recurrence says that

3t3

+ ;((,32)2W2 + (23,)2tzt,) +

;((,

;

,)tf)

3t3

+ 6tzt, +

t;.

3 I

The recurrence for

looks formidable at first, possibly even frightening;

but it really isn’t bad, only convoluted. We can define

= nt,

and then everything simplifies considerably:

uk,

ukj

m>O

k,!

k2!

“’

k,!

’

ifn>l.

(7.82)

kl+kJ+...+k,=n-1

The inner sum is the coefficient of

z+’

.m the egf 0 (z) , raised to the mth

power; and we obtain the correct formula also when n =

if we add in the

term fi(z)O that corresponds to the case m = 0. So

[P’]

;

ti(p

= [z”-‘]

[zn]

,,w

In>0

for all n > 0, and we have the equation

(7.83)

Progress! Equation (7,83) is almost like

E(z) = erEcri,

which defines the generalized exponential series E(z) = El (z) in (5.59) and

(7.70); indeed, we have

cl(z)

z&(z)

So we can read off the answer to our problem:

[zn]

Cl(z) = (n-l)! [z”~‘] E(z) =

nnp2

(7.84)

The complete graph on

,2, . . . , n} has exactly nn ’ spanning trees, for all

n > 0.

356 GENERATING FUNCTIONS

7.7

DIRICHLET GENERATING FUNCTIONS

There are many other possible ways to generate a sequence from a

series; any system of “kernel” functions K,(z) such that

g,,

K,(z) = 0

==+ g,,

= 0 for all n

can be used, at least in principle. Ordinary generating functions use K,(z) =

zn,

and exponential generating functions use K, (z) = 2*/n!; we could also try

falling factorial powers zc, 01: binomial coefficients zs/n! =

(R)

The most important alternative to gf’s and egf’s uses the kernel functions

1 /n”; it is intended for sequences

(41

, 92, .

. . ) that begin with n = 1 instead

of n = 0:

This is called a Dirichlet generating function (dgf), because the German

mathematician Gustav Lejeune Dirichlet (1805-1859) made much of it.

For example, the dgf of the constant sequence (1 , 1 ,

. . . ) is

(7.86)

This is Riemann’s zeta function, which we have also called the generalized

harmonic number Hk’ when

> 1.

The product of Dirichlet generating functions corresponds to a special

kind of convolution:

Thus F(z)

c(z)

= H(z) is the dgf of the sequence

h/d.

(7.87)

d\n

For example, we know from (4.55) that

td,n

p(d) =

[n=

this is

the Dirichlet convolution of the Mobius sequence

(u(

1) ,

2))

3)) . . . ) with

(l,l,l,...),

hence

(7.88)

In other words, the dgf of

(p(l),

FL(~),

p(3), . . .) is

Lo’.

7.7 DIRICHLET GENERATING FUNCTIONS 357

Dirichlet generating functions are particularly valuable when the se-

quence

(gl,g2,...)

is a multiplicative function, namely when

gmn

for m

In such cases the

v,alues

for all n are determined by the values of

g,,

when

n is a power of a prime, and we can factor the dgf into a product over primes:

G(z)

I-I

prime

(

,+!E+w+!!!?L+...

PLZ P3=

If, for instance, we set

= 1 for all n, we obtain a product representation

of Riemann’s zeta function:

L(z)

gm,.(

’

The Mobius function has v(p) = -1 and p(pk) = 0 for k > 1, hence its dgf is

G(z)

-p-“);

(7.91)

prime

this agrees, of course, with (7.88) and (7.90). Euler’s

function has

cp(pk)

Pk-P

k-’ hence its dgf has the factored form

TNe

conclude that

g(z)

I(z

l)/<(z).

Exercises

Warmups

An eccentric collector of 2 x n domino tilings pays $4 for each vertical

domino and $1 for each horizontal domino. How many tilings are worth

exactly $m by this criterion? For example, when m = 6 there are three

solutions:

El, and

Give the generating function and the exponential generating function for

the sequence

(2,5,13,35,.

. . ) = (2” +

3n)

in closed form.

3 What is

~.n~cJ

H,/lOn?

4 The general expansion theorem for rational functions

P(z)/Q(z)

is not

completely general, because it restricts the degree of P to be less than

the degree of Q. What happens if P has a larger degree than this?

358 GENERATING FUNCTIONS

5 Find a generating function S(z) such that

[z”l

S(z)

(;)

(,I,,) .

Basics

6 Show that the recurrence (7.32) can be solved by the repertoire method,

without using generating functions.

7 Solve the recurrence

+29,-2+...+ng0,

for n > 0.

8 What is

[z”]

(ln(1

z))z:‘(l

z)~+‘?

9 Use the result of the previous exercise to evaluate

xE=,

HkHnpk.

Set

= s =

-l/2

in identity (7.61) and then remove all occurrences of

l/2

by using tricks like (5.36). What amazing identity do you deduce?

I deduce that Clark

11 This problem, whose three parts are independent, gives practice in the

Kent is really

superman.

manipulation of generating functions. We assume that A(z) =

x:,

anzn,

B(z) =

bnzn, C(z) =

tncnzn,

and that the coefficients are zero for

negative n.

‘TX

tj+,Zk<n

ojbk, express C in terms of A and B.

If nb, = LET0

2kak/(n

k)!, express A in terms of B.

is a real number and if a, =

IL=,

(‘+kk)bnpk,

express A in

terms of B; then use your formula to find coefficients fk(r) such that

x;=,

fk(T)

an-

12 How many ways are there to put the numbers

,2,.

. . ,2n} into a 2 x n

array so that rows and columns are in increasing order from left to right

and from top to bottom? For example, one solution when n = 5 is

(

12458

3 6 7 910

13 Prove Raney’s generalized lemma, which is stated just before

(7.6~).

14 Solve the recurrence

= 1,

gkgn-k,

for n >

by using an exponential generating function.

7 EXERCISES 359

15 The Bell number

is the number of ways to partition n things into

subsets. For example,

= 5 because we can partition

,2,3}

in the

following ways:

Prove that b,+l =

x.k

(L)bnpk,

and use this recurrence to find a closed

form for the exponential generating function

I,,

b,z”/n!.

Two sequences (a,,) and

(b,,)

are related by the convolution formula

k,i-Zkz+...nk,=n

(al+il-1)

((12+:-l)

,.,

(an+:-‘)

;

also as = 0 a:nd

= 1. Prove that the corresponding generating func-

tions satisfy l:nB(z)

=A(z)

+ iA(z3)

+....

Show that the exponential generating function

G(z)

of a sequence is re-

lated to the ordinary generating function

G(z)

by the formula

G(zt)e-‘dt = G(z),

if the integral exists.

Find the Dirichlet generating functions for the sequences

sn=@;

g,,

= Inn.;

= [n is squarefree].

Express your answers in terms of the zeta function. (Squarefreeness is

defined in exercise 4.13.)

19 Every power series

F(z)

naO

f,z” with

= 1 defines a sequence of

polynomials f,,(x) by the rule

F(z)'

= ~f,(x)z",

II>0

where

f,(

1) =

and

f,(O)

01.

In general, f,(x) has degree n.

Show that such polynomials always satisfy the convolution formulas

fk(X)fn-k(Y)

fn(x

+Y)

;

k=O

(x+Y)kkfk(x)fnpk(Y)

Xnf,(X+y).

kzo

(The identities in Tables 202 and 258 are special cases of this trick.)

360 GENERATING FUNCTIONS

20 A power series G(z) is called differentiably finite if there exist finitely

many polynomials

(z), . . . ,

P,(z), not all zero, such that

Po(z)G(z)+P,(z)G’(z)+-~+P,(z)G(m)(z)

= 0.

A sequence of numbers (go,

,g2,. . . ) is called polynomially recursive

if there exist finitely many polynomials po (z), . . , p,,,(z), not all zero,

such that

Po(n)gn+m(n)gn+l

+...+h(n)~~+~

= 0

for all integers n 3 0. Prove that a generating function is differentiably

finite if and only if its sequence of coefficients is polynomially recursive.

Homework exercises

21 A robber holds up a bank and demands $500 in tens and twenties. He

also demands to know the number of ways in which the cashier can give

him the money. Find a generating function G(z) for which this number

Will

he settle

for

[z500]

G(z), and a more compact generating function G(z) for which

domino

this number is

[z50]

(2).

Determine the required number of ways by

tilings?

(a) using partial fractions; (b) using a method like (7.39).

22 Let P be the sum of all ways to “triangulate” polygons:

(The first term represents a degenerate polygon with only two vertices;

every other term shows a polygon that has been divided into triangles.

For example, a pentagon can be triangulated in five ways.) Define a

“multiplication” operation AAB on triangulated polygons A and B so

that the equation

+ PAP

is valid. Then replace each triangle by

‘z’;

what does this tell you about

the number of ways to decompose an n-gon into triangles?

23 In how many ways can a 2 x 2 x n pillar be built out of 2 x 1 x 1 bricks?

At union rates, as

24 How many spanning trees are in an n-wheel (a graph with n “outer”

many as you can

afford,

plus a few.

vertices in a cycle, each connected to an (n +

1)st

“hub” vertex), when

n 3 3?

7 EXERCISES 361

25 Let m 3 2 be an integer. What is a closed form for the generating

function of the sequence (n mod m), as a function of

and m? Use

this generating function to express ‘n mod m’ in terms of the complex

number w =

eilniirn.

(For example, when m = 2 we have w = -1 and

nmod2=

-5(-l)“.)

26 The second-order Fibonacci numbers

(5,)

are defined by the recurrence

= 5n-I +

for n

Express

in terms of the usual Fibonacci numbers

and F,+r .

27 A 2 x n domino tiling can also be regarded as a way to draw n disjoint

lines in a 2 x n array of points:

If we superimpose two such patterns, we get a set of cycles, since ev-

ery point is touched by two lines. For example, if the lines above are

combined with

,the

lines

the result is

The same set of cycles is also obtained by combining

I I

with

--’

But we get a unique way to reconstruct the original patterns from the

superimposed ones if we assign orientations to the vertical lines by using

arrows that go alternately up/down/up/down/. . . in the first pattern and

alternately down/up/down/up/. . in the second. For example,

The number of such oriented cycle patterns must therefore be

Fi,,

and we should be able to prove this via algebra. Let Q,, be the number

of oriented 2 x n cycle patterns. Find a recurrence for Qn, solve it with

generating functions, and deduce algebraically that Qn =

Fi,,

28 The coefficients of A(z) in (7.89) satisfy A,+A,+ro+A,+20+Ar+30 = 100

for 0 <

< 10. Find a “simple” explanation for this.

362 GENERATING FUNCTIONS

29 What is the sum of Fibonacci products

m>O

+k>+...+k,=n

,kz....,k,>O

30 If the generating function G(z) =

l/(

1x2)(1

(3~)

has the partial

fraction decomposition a/( 1 -KZ) +b/( 1

(3z),

what is the partial fraction

decomposition of G(z)“?

31 What function g(n) of the positive

~integer

n satisfies the recurrence

g(d) cp(n/d) = 1,

d\n

where

is Euler’s totient function?

32 An arithmetic progression is an infinite set of integers

{an+b}

{b,a+b,2a+b,3a+b

,...

A set of arithmetic progressions

{al

n + bl}, . . . , {amn + b,} is called an

exact cover if every nonnegative integer occurs in one and only one of the

progressions. For example, the three progressions

{2n},

{4n + l}, (4n + 3)

constitute an exact cover. Show that if

{al

n +

br},

. . , {amn + b,} is an

exact cover such that 2 6

6 .. . < a,,,, then

a,-1

= a,. Hint: Use

generating functions.

Exam problems

33 What is [w”zn] (ln(1 +

z))/(l

wz)?

34 Find a closed form for the generating function

tn30

Gn(z)wn, if

(Here m is a fixed positive integer.)

35 Evaluate the sum

xO<k,n

/k(n

k) in two ways:

Expand the summand in partial fractions.

b Treat the sum as a convolution and use generating functions.

36 Let A(z) be the generating function for (ac,

al,

as,

. . . ). Express

aln/,,,Jzn in terms of A, z, and m.

7 EXERCISES 363

37 Let a,, be the number of ways to write the positive integer n as a sum of

powers of 2, disregarding order. For example,

= 4, since 4 = 2 + 2 =

2+1+1

=l+l+l+l.

Byconventionweletao=l.

Letb,=tLZoak

be the cumulative sum of the first a’s.

Make a table of the a’s and b’s up through n = 10. What amazing

relation do you observe in your table? (Don’t prove it yet.)

b Express the generating function A(z) as an infinite product.

Use the expression from part (b) to prove the result of part (a).

38 Find a closed form for the double generating function

M(w,z)

min(m,n)w”‘z”

Tll.n30

Generalize your answer to obtain, for fixed m 3 2, a closed form for

M(zI,

..,.z,)

min(n.1,. . , n,)

2:‘.

. .

z”,m

,...,n,30

39 Given positive integers m and n, find closed forms for

k,kz...k,

and

k,kz...k,.

l<k,<kz<...-:k,<n lik,~k>$...<k,,,$n

(For example, when m = 2 and n = 3 the sums are 1.2 +

1.3

2.3

and

+1.2+1.3+2:.2+2.3+3.3.)

Hint: What are the coefficients of

z”’

in the

generating functions (1 + al z) . . (1 +

a,z)

and

l/(

al z) . . . (1

a,z)?

40 Express

(L)(kFk-r

Fk)(n

k)i in closed form.

41 An up-down permutation of order n is an arrangement al

. . . a,, of

the integers {1,2,. . . ,n} that goes alternately up and down:

> . ‘.

For example, 35142 is an up-down permutation of order 5. If A, de-

notes the number of up-down permutations of order n, show that the

exponential

gen.erating

function of (A,,) is (1 + sin z)/cos z.

42 A space probe has discovered that organic material on Mars has DNA

composed of five symbols, denoted by (a, b, c, d, e), instead of the four

components in earthling DNA. The four pairs cd, ce, ed, and ee never

occur consecutively in a string of Martian DNA, but any string with-

out forbidden pairs is possible. (Thus bbcda is forbidden but bbdca is

OK.) How marry Martian DNA strings of length n are possible? (When

n = 2 the answer is 21, because the left and right ends of a string are

distinguishable.)

364 GENERATING FUNCTIONS

43 The Newtonian generating function of a sequence

(gn)

is defined to be

Find a convolution formula that defines the relation between sequences

(fn), (gn), and (h,) whose Newtonian generating functions are related

by the equation

i(z)6

(z) =

h(z).

Try to make your formula as simple

and symmetric as possible.

44 Let q,, be the number of possible outcomes when n numbers

{xl,.

,x,}

are compared with each other. For example,

= 13 because the possi-

bilities are

<x2

<x3

;

<:X2 =X3

;

<x3

<x2;

=Xz<Xj;

=X2=X3;

=‘Xj<X2;

<Xl

<x3

;

X2<Xl

=x3; X2 <: x3

;

X2=X3

<Xl

;

Xj<Xl

<x2;

x3<:x1

=x2;

<x2

<Xl

Find a closed form for the egf

o(z)

qnzn/n!. Also find sequences

(a,),

(W,

such

that

tk”ak

k>O

{;}bi;

;(;)ck,

foralln>O.

45 Evaluate

,YYm,n>O

nl/m2n2.

46 Evaluate

in closed form. Hint:

2.3

+ & =

(z+

f)(z-

5)‘.

47 Show that the numbers

and V,, of 3 x n domino tilings, as given in

(7.34), are closely related to the fractions in the Stern-Brocot tree that

converge to

48 A certain sequence

(gn:)

satisfies the recurrence

ag, + bg,+l +

cgrr+2

integer n 3 0,

for some integers (a, b, c, d) with gcd(a, b, c, d) = 1. It also has the closed

form

[c~(

1 +

Jz)“]

, integer n 3 0,

for some real number

between 0 and 1. Find a, b, c, d, and a.

‘7 EXERCISES 365

Kissinger, take note.

49 This is a problem about powers and parity.

Consider the sequence

(ao,

al,

a2,. . . ) = (2,2,6,. . . ) defined by the

formula

da"

l/2)".

Find a

sim:ple

recurrence relation that is satisfied by this sequence.

b Prove that

[(l

+ &!)“I

n (mod 2) for all integers n > 0.

C Find a number

of the form (p +

$7)/2,

where p and q are positive

integers, such that

LLX”]

n (mod 2) for all integers n > 0.

Bonus problems

50 Continuing exercise 22, consider the sum of all ways to decompose poly-

gons into polygons:

Q=-tA+n++++

+(>+(p+gJ+ft+~+Q+Q+...

Find a symbolic equation for Q and use it to find a generating function

for the number of ways to draw nonintersecting diagonals inside a convex

n-gon. (Give a closed form for the generating function as a function of

you need not

find

a closed form for the coefficients.)

51 Prove that the product

cos

mfl

this a hint or a

warning?

is the generating function for tilings of an m x n rectangle with dominoes.

(There are mn factors, which we can imagine are written in the mn cells

of the rectangle. If mn is odd, the middle factor is zero. The coefficient

II~

ok is the number of ways to do the tiling with j vertical and k

horizontal dominoes.) Hint: This is a difficult problem, really beyond

the scope of this book. You may wish to simply verify the formula in the

case m = 3, n

= 4.

Prove that the polynomials defined by the recurrence

P*(Y)

;)”

(;)

(;)n-kpkh),

integer n 3 0,

have the form p,,(y) = x.“,=, IcIy”, where Ii1 is a positive integer for

1 6 m 6 n. Hint: This exercise is very instructive but not very easy.

366 GENERATING FUNCTIONS

53 The sequence of pentagonal numbers

(1,5,12,22,.

. . ) generalizes the

triangular and square numbers in an obvious way:

Let the nth triangular number be

T,,

n(n+1)/2;

let the nth pentagonal

number be

n(3n

1)/2; and let

Ll,,

be the 3 x n domino-tiling

number defined in (7.38). Prove that the triangular number

TIuq,+Lml

i,z

is also a pentagonal number. Hint:

3Ui,

q =

(Vznml

+ Vln+l)’ + 2.

54 Consider the following curious construction:

1 2 3 4

5 6 7 8 9 10

12 13 14 15 16

. . .

12 3 4

6 7 8 9

12 13 14 16

. . .

1 3 610

63 76 90 106

. . .

31 51

63 76 106

. . .

1 4

26 49 80

131 194

270 376

. . .

26 49

131

194 376

. . .

211

405

781

. . .

31 211 781

. . .

32 243 1024

. . .

(Start with a row containing all the positive integers. Then delete every

mth column; here m = 5. Then replace the remaining entries by partial

sums. Then delete every (m

1 )st column. Then replace with partial

sums again, and so on.) Use generating functions to show that the final

result is the sequence of mth powers. For example, when m = 5 we get

(15,25,35,45

,...)

asshown.

55 Prove that if the power series

F(z)

and G(z) are differentiably finite (as

defined in exercise 20), then so are

F(z)

+ G(z) and

F(z)G(z).

Research problems

56 Prove that there is no “simple closed form” for the coefficient of

Z”

(1 +

+ z~)~, as a function of n, in some large class of “simple closed

forms!’

5’7

Prove or disprove: If all the coefficients of G(z) are either 0 or 1, and if

all the coefficients of G (2)’ are less than some constant M, then infinitely

many of the coefficients of

G(z)’

are zero.

Discrete Probability

THE ELEMENT OF CHANCE enters into many of our attempts to under-

stand the world we live in. A mathematical theory of probability allows us

to calculate the likelihood of complex events if we assume that the events are

governed by appropriate axioms. This theory has significant applications in

all branches of science, and it has strong connections with the techniques we

have studied in previous chapters.

Probabilities are called “discrete” if we can compute the probabilities of

all events by summation instead of by integration. We are getting pretty good

at sums, so it should come as no great surprise that we are ready to apply

our knowledge to some interesting calculations of probabilities and averages.

8.1 DEFINITIONS

(Readers unfamiliar

Probability theory starts with the idea of a probability space, which

with probability

theory will, with

is a set

of all things that can happen in a given problem together with a

high probability,

rule that assigns a probability Pr(w) to each elementary event w

The

benefit from a

probability Pr(w) must be a nonnegative real number, and the condition

perusal of Feller’s

classic introduction

to the subject [96].)

Pr(w) = 1

(8.1)

WEn

must hold in every

dimscrete

probability space. Thus, each value Pr(w) must lie

in the interval

. .

11.

We speak of Pr as a probability distribution, because

it distributes a total probability of 1 among the events w.

Here’s an example: If we’re rolling a pair of dice, the set

of elementary

events is D2 =

{

E],

D, . . . ,

a}, where

Never say die.

is the set of all six ways that a given die can land. Two rolls such as q u

and q n are considered to be distinct; hence this probability space has a

367

368 DISCRETE PROBABILITY

total of

6’

= 36 elements.

We usually assume that dice are “fair,” namely that each of the six possi-

bilities for a particular die has probability

and that each of the 36 possible

rolls in

has probability

But we can also consider “loaded” dice in which

Careful: They

there is a different distribution of probabilities. For example, let

might go off.

Prl(m)

= Pr,(m) =

Prl(a)

Prl(m)

Prj(m)

Prl(m)

Then

LED

Prl

(d) = 1, so

Prl

is a probability distribution on the set D, and

we can assign probabilities to the elements of

by the rule

Pr,,(dd’) = Prl(d) Prl(d’).

(8.2)

For example, Prlj ( q m) =

This is a valid distribution because

Prll(w) =

Prll(dd’)

Prl(d) Prl(d’)

wen

dd’EDZ

d,d’ED

Prl(d)

Prr(d’) = 1 . 1 = 1 .

dED

d’ED

We can also consider the case of one fair die and one loaded die,

Prol(dd’) = Pro(d) Prl(d’),

where Pro(d) =

(8.3)

in which case

ProI

( q m) =

Dice in the “real world” can’t really

be expected to turn up equally often on each side, because there is not perfect

all

sides of

cube

symmetry; but

is usually pretty close to the truth.

were identical,

how

An event is a subset of

In dice games, for example, the set

could we tell which

side is face up?

is the event that “doubles are thrown!’

The individual elements

are

called elementary events because they cannot be decomposed into smaller

subsets; we can think of

as a one-element event {w}.

The probability of an event A is defined by the formula

Pr(wE A) =

Pr(w);

WEA

(8.4)

and in general if R(o) is any statement about

we write

‘Pr(R(w))’

for the

sum of all

Pr(w)

such that R(w) is true. Thus, for example, the probability of

doubles with fair dice is

but when both dice are

loaded with probability distribution

Prl

it is

1+~+~+~+~+~

16 64 64 64 64 16

Loading the dice makes the event “doubles are thrown” more probable.

8.1 DEFINITIONS 369

(We have been using x-notation in a more general sense here than de-

fined in Chapter 2: The sums in (8.1) and (8.4) occur over all elements w

of an arbitrary set, not over integers only. However, this new development is

not really alarming; we can agree to use special notation under a

whenever

nonintegers are intended, so there will be no confusion with our ordinary con-

ventions. The other definitions in Chapter 2 are still valid; in particular, the

definition of infinite

,sums

in that chapter gives the appropriate interpretation

to our sums when the set

is infinite. Each probability is nonnegative, and

the sum of all proba’bilities is bounded, so the probability of event A in (8.4)

is well defined for all subsets A

n.)

A random variable is a function defined on the elementary events w of a

probability space. For example, if

we can define S(w) to be the sum

of the spots on the dice roll w, so that S( q m) = 6 + 3 = 9. The probability

that the spots total seven is the probability of the event S(w) = 7, namely

Pr(Om)

Pr(mm)

Pr(mn)

Pr(flE])

Pr(mn)

Pr(mm)

With fair dice (Pr = Proo), this happens with probability

with loaded dice

(Pr =

Prl,

), it happens with probability & + & + & + & + & +

= &,

the same as we observed for doubles.

It’s customary to drop the ‘(w)’ when we talk about random variables,

because there’s usually only one probability space involved when we’re work-

ing on any particular problem. Thus we say simply ‘S = 7’ for the event that

a 7 was rolled, and ‘S = 4’ for the event

{

A random varialble can be characterized by the probability distribution of

its values. Thus, for example, S takes on eleven possible values {2,3, . . . ,12},

and we can tabulate the probability that S = s for each s in this set:

3 4 5 6 7

9 10

z z z w

w w

If we’re working on a. problem that involves only the random variable S and no

other properties of dice, we can compute the answer from these probabilities

alone, without regard to the details of the set

= D2. In fact, we could

define the probability space to be the smaller set

= {2,3,. . . ,12}, with

whatever probabilikv distribution Pr(s) is desired. Then

‘S

= 4’ would be

an elementary event. Thus we can often ignore the underlying probability

space

and work directly with random variables and their distributions.

If two random variables X and Y are defined over the same probabil-

ity space

we can charactedze their behavior without knowing everything

370 DISCRETE PROBABILITY

about R if we know the

‘joi.nt

distribution”

Just Say

No.

Pr(X=x and

Y=y)

for each x in the range of X and each y in the range of Y. We say that X and

Y are independent random variables if

Pr(X=x and Y=y) =

Pr(X=x).

Pr(Y=y)

(8.5)

for all x and y. Intuitively, this means that the value of X has no effect on

the value of Y.

For example, if

is the set of dice rolls D2, we can let

be the number

of spots on the first die and

the number of spots on the second. Then

the random variables

and

are independent with respect to each of the

probability distributions Prcc,

Prl,

, and

ProI

discussed earlier, because we

defined the dice probability for each elementary event dd’ as a product of a

probability for

= d multiplied by a probability for

= d’. We could have

defined probabilities differently so that, say,

pr(am)

Pr(mm)

Pr(aa)

Pr(Om);

but we didn’t do that, because different dice aren’t supposed to influence each

other. With our definitions, both of these ratios are Pr(S2

=5)/

Pr(S2

=6).

We have defined S to be the sum of the two spot values,

SZ.

Let’s

consider another random variable P, the product SlS2. Are S and P indepen-

dent? Informally, no; if we are told that S = 2, we know that P must be 1.

Formally, no again, because the independence condition (8.5) fails spectac-

ularly (at least in the case of fair dice): For all legal values of s and

have 0 <

Proo[S

=s].Proo[P=p] 6

5.4;

this can’t equal

Proo[S

=sandP=p],

which is a multiple of A.

If we want to understand the typical behavior of a given random vari-

able, we often ask about its “average” value. But the notion of “average”

is ambiguous; people generally speak about three different kinds of averages

when a sequence of numbers is given:

the mean (which is the. sum of all values, divided by the number of

values);

the median (which is the middle value, numerically);

the mode (which is the value that occurs most often).

For example, the mean of

(3,1,4,1,5)

3+1+t+1+5

= 2.8; the median is 3;

the mode is

But probability theorists usually work with random variables instead of

with sequences of numbers, so we want to define the notion of an “average” for

random variables too. Suppose we repeat an experiment over and over again,

dicey

inequality.

8.1 DEFINITIONS 371

making independent trials in such a way that each value of X occurs with

a frequency approximately proportional to its probability. (For example, we

might roll a pair of dice many times, observing the values of S and/or P.) We’d

like to define the average value of a random variable so that such experiments

will usually produce a sequence of numbers whose mean, median, or mode is

approximately the

s,ame

as the mean, median, or mode of X, according to our

definitions.

Here’s how it can be done: The mean of a random real-valued variable X

on a probability space

is defined to be

x.Pr(X=:x)

(8.6)

XEX(cl)

if this potentially infinite sum exists. (Here X(n) stands for the set of all

values that X can assume.) The median of X is defined to be the set of all x

such that

Pr(X6x) 3

and Pr(X3x) 2

(8.7)

And the mode of X is defined to be the set of all x such that

Pr(X=x) 3 Pr(X=x’) for all x’

X(n).

(8.8)

In our dice-throwing example, the mean of S turns out to be 2.

+ 3.

+...

12.

= 7 in distribution

Prcc,

and it also turns out to be 7 in

distribution

Prr

The median and mode both turn out to be

(7)

as well,

in both distributions. So S has the same average under all three definitions.

On the other hand the P in distribution

Pro0

turns out to have a mean value

= 12.25; its median is

{lo},

and its mode is

{6,12}.

The mean of P is

unchanged if we load the dice with distribution Prll , but the median drops

{8}

and the mode becomes

{6}

alone.

Probability theorists have a special name and notation for the mean of a

random variable:

Th.ey

call it the

expected

value, and write

EX =

X(w) Pr(w).

wEn

(8.9)

In our dice-throwing example, this sum has 36 terms (one for each element

!J),

while (8.6) is a sum of only eleven terms. But both sums have the

same value, because they’re both equal to

xPr(w)[x=X(w)]

UJEfl

XEX(Cl)

372 DISCRETE PROBABILITY

The mean of a random variable turns out to be more meaningful in

[get

it:

applications than the other kinds of averages, so we shall largely forget about

On average,

“aver-

medians and modes from now on. We will use the terms “expected value,”

age” means “mean.”

“mean,” and “average” almost interchangeably in the rest of this chapter.

If X and Y are any two random variables defined on the same probability

space, then X + Y is also a random variable on that space. By formula (8.g),

the average of their sum is the sum of their averages:

E(X+Y) =

(X(w) +Y(cu)) Pr(cu) = EX+ EY.

WEfl

(8.10)

Similarly, if

is any constant we have the simple rule

E(oLX)

= REX. (8.11)

But the corresponding rule for multiplication of random variables is more

complicated in general; the expected value is defined as a sum over elementary

events, and sums of products don’t often have a simple form. In spite of this

difficulty, there is a very nice formula for the mean of a product in the special

case that the random variables are independent:

E(XY) = (EX)(EY), if X and Y are independent. (8.12)

We can prove this by the distributive law for products,

E(XY) =

X(w)Y(cu).Pr(w)

WEfl

xy.Pr(X=x

and Y=y)

xcx(n)

YEY(fl)

xy.Pr(X=x)

Pr(Y=y)

?&X(n)

YEY(fl)

xPr(X=x) .

yPr(Y=y)

= (EX)(EY).

XEX(cll

EY(n)

For example, we know that S =

+Sl

and P =

SZ,

when

and

are

the numbers of spots on the first and second of a pair of random dice. We have

ES, =

ES2

= 5, hence ES = 7; furthermore

and

are independent, so

EP =

G.G

as claimedearlier. We also have E(S+P) = ES+EP =

7+7.

But S and P are not independent, so we cannot assert that E(SP) =

7.y

In fact, the expected value of SP turns out to equal

in distribution

Prco,

112 (exactly) in distribution Prlr .

(Slightly subtle

point:

There

are two

probability

spaces,

depending on what

strategy we use;

but

EX,

and

EXz

are

the

same in both.)

8.2 MEAN AND VARIANCE 373

8.2 MEAN AND VARIANCE

The next most important property of a random variable, after we

know its expected value, is its variance, defined as the mean square deviation

from the mean:

?X = E((X

E-X)‘)

(8.13)

If we denote EX by

~1,

the variance VX is the expected value of (X-

FL)‘.

This

measures the “spread” of X’s distribution.

As a simple exa:mple of variance computation, let’s suppose we have just

been made an offer we can’t refuse: Someone has given us two gift certificates

for a certain lottery. The lottery organizers sell 100 tickets for each weekly

drawing. One of these tickets is selected by a uniformly random process-

that is, each ticket is equally likely to be chosen-and the lucky ticket holder

wins a hundred million dollars. The other 99 ticket holders win nothing.

We can use our gift in two ways: Either we buy two tickets in the same

lottery, or we buy

‘one

ticket in each of two lotteries. Which is a better

strategy? Let’s try to analyze this by letting

and

be random variables

that represent the amount we win on our first and second ticket. The expected

value of

X1,

in millions, is

EX, =

~~O+&,.lOO

= 1,

and the same holds for

X2.

Expected values are additive, so our average total

winnings will be

E(X1

X2)

‘EX,

EX2

= 2 million dollars,

regardless of which strategy we adopt.

Still, the two strategies seem different. Let’s look beyond expected values

and study the exact probability distribution of

X2:

winnings (millions)

100

200

same drawing

.9800 .0200

different drawings

.9801

.0198

.OOOl

If we buy two tickets in the same lottery we have a 98% chance of winning

nothing and a 2% chance of winning $100 million. If we buy them in different

lotteries we have a 98.01% chance of winning nothing, so this is slightly more

likely than before; a.nd we have a 0.01% chance of winning $200 million, also

slightly more likely than before; and our chances of winning $100 million are

now 1.98%. So the distribution of

in this second situation is slightly

374 DISCRETE PROBABILITY

more spread out; the middle value, $100 million, is slightly less likely, but the

extreme values are slightly more likely.

It’s this notion of the spread of a random variable that the variance is

intended to capture. We measure the spread in terms of the squared deviation

of the random variable from its mean. In case 1, the variance is therefore

.SS(OM

2M)’ + .02(

1OOM

2M)’ =

196M2

;

in case 2 it is

.9801

(OM

2M)’ + .0198( 1 OOM

2M)2 + .0001(200M

2M)’

= 198M2.

As we expected, the latter variance is slightly larger, because the distribution

of case 2 is slightly more spread out.

When we work with variances, everything is squared, so the numbers can

get pretty big. (The factor

is one trillion, which is somewhat imposing

Interesting: The

even for high-stakes gamblers.) To convert the numbers back to the more

variance of a dollar

meaningful original scale, we often take the square root of the variance. The

amount is expressed

in units of square

resulting number is called the standard deviation, and it is usually denoted

dollars.

by the Greek letter

0=&Z.

(8.14)

The standard deviations of the random variables X’ +

in our two lottery

strategies are

&%%?

= 14.00M and

&?%?

14.071247M. In some sense

the second alternative is about $71,247 riskier.

How does the variance help us choose a strategy? It’s not clear. The

strategy with higher variance is a little riskier; but do we get the most for our

money by taking more risks or by playing it safe? Suppose we had the chance

to buy 100 tickets instead of only two. Then we could have a guaranteed

victory in a single lottery (and the variance would be zero); or we could

gamble on a hundred different lotteries, with a .99”’

.366 chance of winning

nothing but also with a nonzero probability of winning up to $10,000,000,000.

To decide between these alternatives is beyond the scope of this book; all we

can do here is explain how to do the calculations.

In fact, there is a simpler way to calculate the variance, instead of using

the definition (8.13). (We suspect that there must be something going on

in the mathematics behind the scenes, because the variances in the lottery

example magically came out to be integer multiples of M’.) We have

Another way to

reduce risk might

be to bribe the

lottery oficials.

guess that’s where

probability becomes

indiscreet.

(N.B.: Opinions

expressed in these

margins do not

necessarily represent

the opinions of the

management.)

E((X

EX)‘) = E(X2

ZX(EX)

+ (EX)‘)

= E(X’)

2(EX)(EX) + (EX)’ ,

8.2 MEAN AND VARIANCE 375

since (EX) is a constant; hence

VX =

E(X’)

(EX)‘.

(8.15)

“The variance is the mean of the square minus the square of the mean.”

For example, the mean of

(Xl

+X2)’ comes to

.98(0M)2

.02(

100M)2

200M’ or to

.9801

I(OM)2

.0198(

100M)’

.OOOl

(200M)2

202M2

in the

lottery problem. Subtracting

4M2

(the square of the mean) gives the results

we obtained the hard way.

There’s an even easier formula yet, if we want to calculate V(X+ Y) when

X and Y are independent: We have

E((X+Y)‘) =

E(X2

+2XY+Yz)

E(X’)

+2(EX)(EY) + E(Y’),

since we know that E(XY) = (EX) (EY) in the independent case. Therefore

V(X + Y) =

E#((X

Y)‘)

(EX +

EY)’

EI:X’)

Z(EX)(EY)

E(Y’)

-- (EX)‘-2(EX)(EY)

(EY)’

El:X’)

(EX)’

E(Y’)

(EY)’

VxtvY.

(8.16)

“The variance of a sum of independent random variables is the sum of their

variances.” For example, the variance of the amount we can win with a single

lottery ticket is

E(X:)

(EXl

)’

.99(0M)2

.Ol(lOOM)’

(1 M)’ =

99M2

Therefore the variance of the total winnings of two lottery tickets in two

separate (independent) lotteries is 2x

99M2

198M2.

And the corresponding

variance for n independent lottery tickets is n x

99M2.

The variance of the dice-roll sum S drops out of this same formula, since

S =

is the sum of two independent random variables. We have

;(12+22+32+42+52+62!-

;

when the dice are fair; hence VS =

The loaded die has

VSI

;(2.12+22+32+42+52+2.62)-

376 DISCRETE PROBABILITY

hence VS =

= 7.5 when both dice are loaded. Notice that the loaded dice

give S a larger variance, although S actually assumes its average value 7 more

often than it would with fair dice. If our goal is to shoot lots of lucky

7’s,

the

variance is not our best indicator of success.

OK, we have learned how to compute variances. But we haven’t really

seen a good reason why the variance is a natural thing to compute. Everybody

does it, but why? The main reason is Chebyshew’s inequality ([24’] and

he proved it in

[50’]), which states that the variance has a significant property:

1867,

it’s a classic

‘67 Chebyshev.

Pr((X-EX)‘>a)

< VX/ol, for all a > 0.

(8.17)

(This is different from the summation inequalities of Chebyshev that we en-

countered in Chapter 2.) Very roughly, (8.17) tells us that a random variable X

will rarely be far from its mean EX if its variance VX is small. The proof is

amazingly simple. We have

VX =

(X(w)

EX:? Pr(w)

CLJE~~

(X(w)

-EXf

Pr(cu)

WEn

(X(w)-EX)‘>a

aPr(w)

= oL.Pr((X

EX)’ > a)

;

WEn

(X(W)-EX]~&~

dividing by a finishes the proof.

If we write u for the mean and o for the standard deviation, and if we

replace

by c2VX in (8.17), the condition (X

EX)’ 3 c2VX is the same as

FL)

(~0)~;

hence (8.17) says that

Pr(/X

~13 co) 6

l/c’.

(8.18)

Thus, X will lie within c standard deviations of its mean value except with

probability at most

l/c’.

A random variable will lie within 20 of

at least

75% of the time; it will lie between u

100 and

+ 100 at least 99% of the

time. These are the cases

4VX and

1OOVX

of Chebyshev’s inequality.

If we roll a pair of fair dice n times, the total value of the n rolls will

almost always be near 7n, for large n. Here’s why: The variance of n in-

dependent rolls is

Fn.

A variance of

means a standard deviation of

only

(That is, the aver-

age will fall between

the stated limits in

at least 99% of all

cases when we look

at a set of n inde-

pendent samples,

for any fixed value

of n Don’t mis-

understand this as

a statement about

the averages of an

infinite sequence

Xl,

x2, x3, .

as n varies.)

8.2 MEAN AND VARIANCE 377

So Chebyshev’s inequality tells us that the final sum will lie between

7n-lO@

and

7n+lO@

in at least 99% of all experiments when n fair dice are rolled. For example,

the odds are better than 99 to 1 that the total value of a million rolls will be

between 6.976 million and 7.024 million.

In general, let X be any random variable over a probability space

hav-

ing finite mean p and finite standard deviation

Then we can consider the

probability space

0”

whose elementary events are n-tuples

(WI,

~2,.

. . ,

w,)

with each uk

fl,

amd

whose probabilities are

Pr(wl,

~2,.

. . , (u,) =

Pr(wl)

Pr(w2).

. . Pr(cu,) .

If we now define random variables

by the formula

Xk(ul,WZ,...

,%)

x(wk),

the quantity

. . +

is a sum of n independent random variables, which corresponds to taking n

independent “samples” of X on

and adding them together. The mean of

+X2+.

.+X,

ntp,

and the standard deviation is

hence the average

of the n samples,

A(X,

+Xz+..,+X,),

will lie between p

100/J;;

and p + loo/,/K at least 99% of the time. In

other words, if we

dhoose

a large enough value of n, the average of n inde-

pendent samples will almost always be very near the expected value EX. (An

even stronger theorem called the Strong Law of Large Numbers is proved in

textbooks of probability theory; but the simple consequence of Chebyshev’s

inequality that we

h,ave

just derived is enough for our purposes.)

Sometimes we don’t know the characteristics of a probability space, and

we want to estimate the mean of a random variable X by sampling its value

repeatedly. (For exa.mple, we might want to know the average temperature

at noon on a January day in San Francisco; or we may wish to know the

mean life expectancy of insurance agents.) If we have obtained independent

empirical observations

X1,

X2,

. . . ,

X,,

we can guess that the true mean is

approximately

Xl+Xzt".+X,

(8.19)

378 DISCRETE PROBABILITY

And we can also make an estimate of the variance, using the formula

\ix

;y’n

(X,

‘.

X,)2

n-l

n(n-1)

(8.20)

The (n ~ 1)

‘s

in this formula look like typographic errors; it seems they should

be n’s, as in (8.1g), because the true variance VX is defined by expected values

in (8.15). Yet we get a better estimate with n

1 instead of n here, because

definition (8.20) implies that

E(i/X) = VX.

Here’s why:

E(\;/X)

&E(

tx:

k=l

n-l

W2)

k=l

(E(Xi’lj#kl+

E(X')Lj=kl))

j=l

k=l

&(nE(X’)

k(nE(X’)

+n(n-

l)E(X)'))

(8.21)

;

xjxk)

j=l

k=l

E(X')-E(X)“

= VX

(This derivation uses the independence of the observations when it replaces

E(XjXk) by (EX)‘[j

fk]

+ E(X’)[j

=k].)

In practice, experimental results about a random variable X are usually

obtained by calculating a sample mean

and a sample standard de-

viation

fi,

and presenting the answer in the form

‘

b/,/i?

‘.

For

example, here are ten rolls of two supposedly fair dice:

The sample mean of the spot sum S is

fi =

(7+11+8+5+4+6+10+8+8+7)/10

= 7.4;

the sample variance is

(72+112+82+52+42+62+102+82+82+72-10~2)/9

2.12

8.2 MEAN AND VARIANCE 379

We estimate the average spot sum of these dice to be

7.4&2.1/m

= 7.4~tO.7,

on the basis of these experiments.

Let’s work one more example of means and variances, in order to show

how they can be ca.lculated theoretically instead of empirically. One of the

questions we considered in Chapter 5 was the “football victory problem,’

where n hats are thrown into the air and the result is a random permutation

of hats. We showed

fin

equation (5.51) that there’s a probability of ni/n!

that nobody gets

thle

right hat back. We also derived the formula

P(n,k)

’

‘n

(n-k)i

-!&$

for the probability that exactly k people end up with their own hats.

Restating these results in the formalism just learned, we can consider the

probability space

FF,

of all n! permutations n of {1,2,. . . , n}, where Pr(n) =

1 /n! for all n

Fin.

The random variable

Not to be confused

F,(x)

= number of “fixed points” of n , for

Fl,,

with a Fibonacci

number.

measures the number of correct hat-falls in the football victory problem.

Equation (8.22) gives

Pr(F,

= k), but let’s pretend that we don’t know any

such formula; we merely want to study the average value of

F,,

and its stan-

dard deviation.

The average value is, in fact, extremely easy to calculate, avoiding all the

complexities of Cha.pter 5. We simply observe that

F,(n)

F,,I

(7~)

F,,2(74

F,,,(d)

Fn,k(~) = [position k of

is a fixed point] , for n

Fl,.

Hence

EF, = EF,,, i- EF,,z + . . . + EF,,,,

And the expected value of

Fn,k

is simply the probability that

Fn,k

which

is l/n because exactly (n

l)! of the n! permutations n =

~1~2

. . . n,

FF,

have

= k. Therefore

EF, = n/n

1 , for n > 0.

(8.23)

One the average.

On the average, one hat will be in its correct place. “A random permutation

has one fixed point, on the average.”

Now what’s the standard deviation? This question is more difficult, be-

cause the

Fn,k

‘s

are not independent of each other. But we can calculate the

380 DISCRETE PROBABILITY

variance by analyzing the mutual dependencies among them:

E(FL,)

(

fFn,k)i’)

Fn,j

Fn,k)

k=l

j=l

k=l

E(Fn,jl’n,k)

E(Fi,k)+2

E(Fn,j

Fn,k)

j=l

k=l

<k<n

l<j<k<n

(We used a similar trick when we derived (2.33) in Chapter 2.) Now Ft

Fn,k,

Since

Fn,k

is either 0 or 1; hence E(Fi,,) =

EF,,k

= l/n as before. And

if j < k we have

E(F,,j

F,,k)

Pr(rr

has both j and k as fixed points) =

2)!/n! =

l/n(n

1). Therefore

E(FfJ

;

n(n-1)

for n 3 2.

(8.24)

(As a check when n = 3, we have

f02

il’

+ i22 + i32 = 2.) The variance

E(Fi)

(EF,)'

so the standard deviation (like the mean) is 1. “A

random permutation of n 3 2 elements has 1

1 fixed points.”

8.3 PROBABILITY GENERATING FUNCTIONS

If X is a random

varia.ble

that takes only nonnegative integer values,

we can capture its probability distribution nicely by using the techniques of

Chapter 7. The probability generating function or pgf of X is

Gx(z)

~Pr(X=k)zk.

k>O

(8.25)

This power series in

contains all the information about the random vari-

able X. We can also express it in two other ways:

Gx(z)

Pr(w)zX(W)

= E(z’).

WEfl

(8.26)

The coefficients of

Gx(z)

are nonnegative, and they sum to 1; the latter

condition can be written

Gx(1)

= 1.

(8.27)

Conversely, any power series

G(z)

with nonnegative coefficients and with

(1)

is the pgf of some random variable.

8.3 PROBABILITY GENERATING FUNCTIONS 381

The nicest

thin,g

about pgf’s is that they usually simplify the computation

of means and variances. For example, the mean is easily expressed:

EX =

xk.P:r(X=k)

k>O

~Pr(X=k).kzk~‘lr=,

k>O

G;(l).

(8.28)

We simply differentiate the pgf with respect to

and set z = 1.

The variance is only slightly more complicated:

E(X’)

xk*.Pr(X=k)

k>O

xPr(X=k).(k(k-

1)~~~’

+ kzk-‘)

I==,

G;(l)

G;(l).

k>O

Therefore

G;(l)

G&(l)-

G;(l)2.

(8.29)

Equations (8.28) and (8.29) tell us that we can compute the mean and variance

if we can compute the values of two derivatives,

GI,

(1)

and

(1).

We don’t

have to know a closed form for the probabilities; we don’t even have to know

a closed form for

G;c

(z) itself.

It is convenient’ to write

Mean(G) = G'(l),

(8.30)

Var(G)

= G"(l)+ G'(l)-

G'(l)',

(8.31)

when G is any function, since we frequently want to compute these combina-

tions of derivatives.

The second-nicest thing about pgf’s is that they are comparatively sim-

ple functions of

in many important cases. For example, let’s look at the

uniform distribution of order n, in which the random variable takes on each

of the values {0, 1, . ,, . , n

with probability l/n. The pgf in this case is

U,(z)

= ;(l-tz+...+znp')

k&g,

for n

(8.32)

We have a closed form for U,(z) because this is a geometric series.

But this closed form proves to be somewhat embarrassing: When we plug

= 1 (the value of

that’s most critical for the pgf), we get the undefined

ratio O/O, even though U,(z) is a polynomial that is perfectly well defined

at any value of

The value

(1) = 1 is obvious from the non-closed form

382 DISCRETE PROBABILITY

+z+...

+ znP1)/n, yet it seems that we must resort to L’Hospital’s rule

to find

lim,,,

U,(z) if we want to determine

U,(

1) from the closed form.

The determination of

UA(

1) by L’Hospital’s rule will be even harder, because

there will be a factor of (z- 1

1’

in the denominator;

l-l:

(1) will be harder still.

Luckily there’s a nice way out of this dilemma. If G(z) = Ena0 gnzn is

any power series that converges for at least one value of

with

Iz/

> 1, the

power series G’(z) = j-n>OngnznP’ will also have this property, and so will

G”(z), G”‘(z), etc. There/fore by Taylor’s theorem we can write

G(,+t)

G(,)+~~t+~t2+~t3+...;

(8.33)

all derivatives of G(z) at z

1 will appear as coefficients, when G( 1 + t) is

expanded in powers of t.

For example, the derivatives of the uniform pgf U,(z) are easily found

in this way:

(l+t)“-1

U,(l +t) =

;

= k(y)

+;;(;)t+;(;)t2+...+;(;)tn-l

Comparing this to (8.33) gives

U,(l)

= 1;

u;(l)

(n-l)(n-2);

(8.34)

and in general Uim’ (1) = (n -- 1

)“/

(m + 1

although we need only the cases

m = 1 and m = 2 to compute the mean and the variance. The mean of the

uniform distribution is

n-l

ulm

2’

and the variance is

(8.35)

U::(l)+U:,(l)-U:,(l)2

(n-

l)(n-2)

+6(n-l)

(n-l)2

12 12 12

The third-nicest thing about pgf’s is that the product of pgf’s corresponds

to the sum of independent random variables. We learned in Chapters 5 and 7

that the product of generating functions corresponds to the convolution of

sequences; but it’s even more important in applications to know that the

convolution of probabilities corresponds to the sum of independent random

8.3 PROBABILITY GENERATING FUNCTIONS 383

variables. Indeed, if X and Y are random variables that take on nothing but

integer values, the probability that X + Y = n is

Pr(X+Y=n)

xPr(X=kandY=n-k).

If X and Y are independent, we now have

Pr(X+Y=n)

tPr(X=k)

Pr(Y=n-k),

a convolution. Therefore-and this is the punch

line-

Gx+Y(z)

Gx(z)

GY(z),

if X and Y are independent.

(8.37)

Earlier this chapter

‘we

observed that V( X + Y) = VX + VY when X and Y are

independent. Let

F(z)

and G(z) be the pgf’s for X and Y, and let H(z) be the

pgf for X + Y. Then

H(z)

F(z)G(z),

and our formulas (8.28) through

(8.31)

for mean and variance tell us that we

must have

Mean(H) = Mean(F) + Mean(G)

;

(8.38)

Var(H) = Var(F)

+Var(G).

(8.39)

These formulas, which are properties of the derivatives Mean(H) = H’( 1) and

Var(H) = H”( 1) + H’( 1)

H’( 1

)2,

aren’t valid for arbitrary function products

H(z) = F(z)G(z); we have

H’(z) = F’(z)G(z) + F(z)G’(z) ,

H”(z) =

F”(z)G(z)

+2F’(z)G’(z)

F(z)G”(z).

But if we set

= 1,

‘we

can see that (8.38) and (8.39) will be valid in general

provided only that

F(1) = G(1) = 1

(8.40)

and that the derivatives exist. The “probabilities” don’t have to be in

for these formulas to hold. We can normalize the functions F(z) and G(z)

by dividing through by F( 1) and G (1) in order to make this condition valid,

whenever F( 1) and G (1) are nonzero.

Mean and variance aren’t the whole story. They are merely two of an

I’//

graduate magna

infinite series of so-c:alled

cumulant

statistics introduced by the Danish as-

cum ulant.

tronomer Thorvald Nicolai Thiele

[288]

in 1903. The first two cumulants

384 DISCRETE PROBABILITY

and ~2 of a random variable are what we have called the mean and the

variance; there also are higher-order cumulants that express more subtle prop-

erties of a distribution. The general formula

G(et)

$t2

$t3

zt4

. . .

(8.41)

defines the cumulants of all orders, when

G(z)

is the pgf of a random variable.

Let’s look at cumulants more closely. If

G(z)

is the pgf for X, we have

G(et)

tPr(X=k)ekt

Pr(X=k)s

k>O

k,m>O

,+CLlt+ClZt2+E++

...

’

(8.42)

where

Pm =

k”‘Pr(X=k) =

E(Xm).

(8.43)

This quantity

is called the “mth moment” of X. We can take exponentials

on both sides of

(8.41),

obtaining another formula for

G(et):

G(e')

= 1

(K,t+;K;+‘+-*)

(K,t+;K2t2+-.)2

. . .

l! 2!

Kit+

;(K2 + K;)t2

f...

Equating coefficients of powers of t leads to a series of formulas

Plr

(8.44)

CL2

-PL:,

(8.45)

3P1

+&:,

(8.46)

-4WcL3

12&2

-3~;

-6p;,

(8.47)

CL5

-5P1P4

+2opfp3

lop2p3

301~1

FL:

60~:~2 + 24~:~

(8.48)

defining the cumulants in terms of the moments. Notice that ~2 is indeed the

variance,

E(X’)

(EX)2,

as claimed.

Equation (8.41) makes it clear that the cumulants defined by the product

“For these higher

F(z) G (z) of two pgf’s will be the sums of the corresponding cumulants of F(z)

ha’f-invariants

and G(z), because logarithms of products are sums. Therefore all cumulants

shall

propose no

of the sum of independent random variables are additive, just as the mean and

special names.

”

Thiele

12881

variance are. This property makes cumulants more important than moments.

8.3 PROBABILITY GENERATING FUNCTIONS 385

If we take a slightly different tack, writing

G(l

+t)

%t+

zt'

$t'

+ ... ,

equation (8.33) tells us that the

K’S

are the “factorial moments”

Gimi(l)

OLm

Pr(X=k)kEzk-“’

lzz,

k20

xkzl?r(X=k)

k>O

E(X”).

(8.49)

It follows that

G(et) = 1 + y+(et

1) +

$(et

1)2

f..’

l+;!(t+ft2+...)+tL(t2+t3+...)+..

= 1

+er.,t+;(OL2+OL,)t2+..~,

and we can express the cumulants in terms of the derivatives

G’ml(l):

011,

(8.50)

Q =

011

c$,

(8.51)

K3 =

013

+ 3Q +

o(1

3cQoL1

24,

(8.52)

This sequence of formulas yields “additive” identities that extend (8.38) and

(8.39) to all the cumulants.

Let’s get back down to earth and apply these ideas to simple examples.

The simplest case

o’f

a random variable is a “random constant,” where X has

a certain fixed value x with probability 1. In this case Gx(z) = zx, and

In Gx(et) = xt; hence the mean is x and all other cumulants are zero. It

follows that the operation of multiplying any pgf by

increases the mean

by x but leaves the variance and all other cumulants unchanged.

How do probability generating functions apply to dice? The distribution

of spots on one fair die has the pgf

z+z2+23+24+25+26

G(z)

zu6(z),

386 DISCRETE PROBABILITY

where

is the pgf for the uniform distribution of order 6. The factor

‘z’

adds 1 to the mean, so the

m’ean

is 3.5 instead of

= 2.5 as given in (8.35);

but an extra

‘z’

does not affect the variance (8.36), which equals

The pgf for total spots on two independent dice is the square of the pgf

for spots on one die,

Gs(z)

z2+2z3+3z4+4z5+5z6+6z7+5z8+4~9+3~10+2~11+Z12

22u&)z.

If we roll a pair of fair dice n times, the probability that we get a total of

k spots overall is, similarly,

[zk]

Gs(z)” =

[zk]

zZnU~;(z)

[zkp2y

u(;

(z)2n

In the hats-off-to-football-victory problem considered earlier, otherwise

Hat distribution is

known as the problem of enumerating the fixed points of a random permuta-

different

kind

tion, we know from (5.49) that the pgf is

uniform distribu-

tion.

F,(z)

(n?!

O<k<n

(n-k)! k!

’

for n 3 0.

(8.53)

Therefore

F,!(z)

k)i Zk-’

,<k<n

(n-k)!

(k-l)!

\..

,<&-,

E3;

. .

F,pl(z).

Without knowing the details of the coefficients, we can conclude from this

recurrence FL(z) =

F,-,(z)

that

F~m’(z)

= F,-,(z); hence

FCml(l) =

F,-,(l)

[n>m].

(8.54)

This formula makes it easy to calculate the mean and variance; we find as

before (but more quickly) that they are both equal to 1 when n 3 2.

In fact, we can now show that the mth cumulant

of this random

variable is equal to 1 whenever n 3 m. For the mth cumulant depends only

FL(l),

F:(l),

. . . .

Fim'(l),

and these are all equal to 1; hence we obtain

8.3 PROBABILITY GENERATING FUNCTIONS 387

Con artists know

that p

0.1

when you spin a

newly minted U.S.

penny

on a

smooth

table. (The weight

distribution makes

Lincoln’s head fall

downward.)

the same answer for the mth cumulant as we do when we replace F,(z) by

the limiting pgf

F,(z) =

e’-’

(8.55)

which has

FE’

( 1)

1 for derivatives of all orders. The cumulants of

are

identically equal to

because

lnF,(et)

= lneet-’ =

8.4 FLIPPING COINS

Now let’s turn to processes that have just two outcomes. If we flip

a coin, there’s probability p that it comes up heads and probability q that it

comes up tails, where

psq

= 1.

(We assume that the coin doesn’t come to rest on its edge, or fall into a hole,

etc.) Throughout this section, the numbers p and q will always sum to 1. If

the coin is fair, we have p = q =

otherwise the coin is said to be biased.

The probability generating function for the number of heads after one

toss of a coin is

H(z) =

q+pz.

(8.56)

If we toss the coin n times, always assuming that different coin tosses are

independent, the number of heads is generated by

H(z)” = (q

+pz)”

(;)pkqn-*zk,

k>O

(8.57)

according to the binomial theorem. Thus, the chance that we obtain exactly k

heads in n tosses is

(i)

p q

This sequence of probabilities is called the

binomial distribution.

Suppose we toss a coin repeatedly until heads first turns up. What is

the probability that exactly k tosses will be required? We have k = 1 with

probability p (since this is the probability of heads on the first flip); we have

k = 2 with probability qp (since this is the probability of tails first, then

heads); and for general k the probability is

qkm’p.

So the generating function

pz+qpz2+q=pz3+-

Gzqz’

(8.58)

388 DISCRETE PROBABILITY

Repeating the process until n heads are obtained gives the pgf

(

)

(n+;-yq,lk

-qz

This, incidentally, is

Z”

times

(&)”

;

(ni-;-l)p.,q’z*.

(8.60)

the generating function for the negative binomial distribution.

The probability space in example

(8.5g),

where we flip a coin until

n heads have appeared, is different from the probability spaces we’ve seen

earlier in this chapter, because it contains infinitely many elements. Each el-

ement is a finite sequence of heads and/or tails, containing precisely n heads

in all, and ending with heads; the probability of such a sequence is

pnqkpn,

Heads

win,

where k

n is the number of tails. Thus, for example, if n = 3 and if we

tails you lose.

write H for heads and T for tails, the sequence THTTTHH is an element of the

No? OK; tails you

probability space, and its probability is qpqqqpp =

p3q4.

lose, heads I win.

Let X be a random variable with the binomial distribution (8.57), and let

No?

Well,

then,

Y be a random variable with the negative binomial distribution (8.60). These

heads

you

,ose

’

tails

win.

distributions depend on n and p. The mean of X is nH’(l) = np, since its

pgf is

Hi;

the variance is

n(H”(1)+H’(1)-H’(1)2)

n(O+p-p2)

= npq.

(8.61)

Thus the standard deviation is

If we toss a coin n times, we expect

to get heads about np

fitpq

times. The mean and variance of Y can be

found in a similar way: If we let

we have

G’(z) =

T9sz,, ,

2pq2

G”(z)

= (, _

qz13

;

hence G’(1) =

pq/p2

= q/p and G”(1) =

2pq2/p3

2q2/p2.

It follows that

the mean of Y is nq/p and the variance is nq/p2.

8.4 FLIPPING COINS 389

A simpler way to derive the mean and variance of Y is to use the reciprocal

generating function

F(z)

l-q2

1 q

---2,

P P

(8.62)

and to write

G(z)” =

F(z)-“.

(8.63)

This polynomial F(z) is not a probability generating function, because it has

a negative coefficient. But it does satisfy the crucial condition F(1) =

Thus

F(z)

is formally a binomial that corresponds to a coin for which we

The probability is

get heads with “probability” equal to -q/p; and G(z) is formally equivalent

negative that I’m

getting younger.

to flipping such a coin

-1 times(!). The negative binomial distribution

with parameters (n,p) can therefore be regarded as the ordinary binomial

Oh? Then it’s

that you’re getting

distribution with parameters (n’, p’) = (-n, -q/p). Proceeding formally,

older, or staying

the mean must be n’p’ = (-n)(-q/p) =

nq/p,

and the variance must be

the same.

n’p’q’ =

(-n)(-q/P)(l

+ 4/p) =

w/p

This formal derivation involving

negative probabilities is valid, because our derivation for ordinary binomials

was based on identities between formal power series in which the assumption

0 6 p 6 1 was never used.

Let’s move on to another example: How many times do we have to flip

a coin until we get heads twice in a row? The probability space now consists

of all sequences of H’s and T's that end with HH but have no consecutive H’s

until the final position:

{HH,THH,TTHH,HTHH,TTTHH,THTHH,HTTHH,.

. .}.

The probability of any given sequence is obtained by replacing H by p and T

by q; for example, the sequence THTHH will occur with probability

Pr(THTHH) = qpqpp = p3q2.

We can now play with generating functions as we did at the beginning

of Chapter 7, letting S be the infinite sum

THH + TTHH + HTHH + TTTHH + THTHH + HTTHH + . . .

of all the elements of fI. If we replace each H by pz and each T by qz, we get

the probability generating function for the number of flips needed until two

consecutive heads turn up.

390 DISCRETE PROBABILITY

There’s a curious relatio:n between S and the sum of domino tilings

in equation (7.1). Indeed, we obtain S from T if we replace each

by T and

each E by HT, then tack on an HH at the end. This correspondence is easy to

prove because each element of

has the form

HT)"HH

for some n 3 0,

and each term of T has the form

+ E)n. Therefore by (7.4) we have

(I-T-HT)-'HH,

and the probability generatin.g function for our problem is

G(z)

-w-

(P~W-‘(PZ)Z

p*2*

= 1

qz-pqz* .

(8.64)

Our experience with the negative binomial distribution gives us a clue

that we can most easily calcmate the mean and variance of (8.64) by writing

where

F(z)

qz-pqz*

’

and by calculating the “mean” and “variance” of this pseudo-pgf F(z). (Once

again we’ve introduced a function with F( 1) = 1.) We have

F’(1) =

(-q-2pq)/p*

2-p-l

-P-*;

F”(1) = -2pq/p* = 2

2pP’

Therefore, since

= F(z)G(z),

Mean

= 2, and Var(z2) = 0, the mean

and variance of distribution G(z) are

Mean(G) = 2

Mean(F) =

pp2

p-l

;

(8.65)

Var(G) = -Va.r(F)

pP4

t&-3

-2~-*-~-1.

(8.66)

When p =

the mean and variance are 6 and 22, respectively. (Exercise 4

discusses the calculation of means and variances by subtraction.)

8.4 FLIPPING COINS 391

”

‘You really are

automaton-a

cal-

culating machine,

’

cried. ‘There is

something

positively

inhuman in you

times.“’

-J.

Watson

(701

Now let’s try a more intricate experiment: We will flip coins until the

pattern THTTH is first obtained. The sum of winning positions is now

S = THTTH

HTHTTH + TTHTTH

HHTHTTH

HTTHTTH + THTHTTH + TTTHTTH + .

;

this sum is more difficult to describe than the previous one. If we go back to

the method by which we solved the domino problems in Chapter 7, we can

obtain a formula for S by considering it as a “finite state language” defined

by the following “automaton”:

The elementary events in the probability space are the sequences of H’s and

T’s that lead from state 0 to state 5. Suppose, for example, that we have

just seen THT; then we are in state 3. Flipping tails now takes us to state 4;

flipping heads in state 3 would take us to state 2 (not all the way back to

state 0, since the TH we’ve just seen may be followed by TTH).

In this formulation, we can let

be the sum of all sequences of H’s and

T’s that lead to state k: it follows that

l+SoH+SzH,

= SoT+S,T+SqT,

H+ S3H,

S2T,

SST,

S5 =

Now the sum S in our problem is

S5;

we can obtain it by solving these six

equations in the six unknowns

SO,

S1,

. . . ,

Sg.

Replacing H by pz and T by qz

gives generating functions where the coefficient of

z”

is the probability

that we are in state k after n flips.

In the same way, any diagram of transitions between states, where the

transition from state j to state k occurs with given probability

pj,k,

leads to

a set of simultaneous linear equations whose solutions are generating func-

tions for the state probabilities after n transitions have occurred. Systems

of this kind are called Markov processes, and the theory of their behavior is

intimately related to the theory of linear equations.

392

DISCRETE PROBABILITY

But the coin-flipping problem can be solved in a much simpler way,

without the complexities of the general finite-state approach. Instead of six

equations in six unknowns

SO,

, . , . ,

Ss,

we can characterize S with only

two equations in two unknowns. The trick is to consider the auxiliary sum

N =

of all flip sequences that don’t contain any occur-

rences of the given pattern THTTH:

N =

+ H + T + HH + . . . + THTHT + THTTT + .

We have

l+N(H+T)

= N+S,

(8.67)

because every term on the left either ends with THTTH (and belongs to S) or

doesn’t (and belongs to N); conversely, every term on the right is either empty

or belongs to N H or N T. And we also have the important additional equation

NTHTTH = S+STTH, (8.68)

because every term on the left completes a term of S after either the first H

or the second H, and because every term on the right belongs to the left.

The solution to these two simultaneous equations is easily obtained: We

have N = (1

S)( 1

’

from

(&X67),

hence

-S)(l

-T-H) ‘THTTH =

S(1

+TTH).

As before, we get the probability generating function G(Z) for the number of

flips if we replace H by

p+q=l,andwefind

G(z))p2q3t5

1-Z

hence the solution is

and T by qz. A bit of simplification occurs since

G(z)(l

+pqV);

G(z)

p2q’;z5

p2q325 + (1

+pqV)(l

(8.69)

Notice that G( 1) = 1, if pq # 0; we do eventually encounter the pattern

THTTH, with probability

unless the coin is rigged so that it always comes

up heads or always tails.

To get the mean and variance of the distribution (8.6g), we invert G(z)

as we did in the previous problem, writing G(z) =

z5/F(z)

where F is a poly-

nomial:

F(z)

p2q3z5+

(1 +pq2z3)(1

p2q3

(8.70)

8.4 FLIPPING COINS 393

The relevant derivatives are

F’(1) = 5

(1 +pq2)/p2q3,

F”(1) = 20

6pq2/p2q3

;

and if X is the number of flips we get

EX = Mean(G) = 5-Mean(F)

pP2qm3

+pmlqP1;

VX = Var(G) =

-Var(F)

(8.71)

-25+pP2q

+ 7p ‘q--l +Mean(F)’

(EX)2

-9~~

2qP”

3pP’qm’

(8.72)

When p =

the mean and variance are 36 and 996.

Let’s get general: The problem we have just solved was “random” enough

to show us how to analyze the case that we are waiting for the first appearance

of an arbitrary pattern A of heads and tails. Again we let S be the sum of

all winning sequences of H's and T’s, and we let N be the sum of all sequences

that haven’t encountered the pattern A yet. Equation (8.67) will remain the

same; equation (8.68) will become

NA =

s(l

+ A”) [A(“-‘,

=A,,_,,]

A(21

[A’m

=A(,-

2,]

+.,.$-Aim

"[A~'-Ac,i]),

(8.73)

where m is the length of A, and where

ACkl

and Aiki denote respectively the

last k characters and the first k characters of A. For example, if A is the

pattern THTTH we just studied, we have

Ai” =

Al21

TH,

Ai31

TTH,

Ai41

HTTH.

A,,, = T,

42, =

TH,

A(3)

= THT, A,,, =

THTT:

Since the only perfect match is

Ai21

= A

,l),

equation (8.73) reduces to (8.68).

Let A be the result of substituting

p-’

for H and

qm’

for T in the pat-

tern A. Then it is not difficult to generalize our derivation of (8.71) and (8.72)

to conclude (exercise 20) that the general mean and variance are

EX =

A/k,

[Alk)

=A/k,]

;

(8.74)

k=l

(EX)2

(2k-

l&k)

[ACk’

=A[k)] .

(8.75)

k=l

394 DISCRETE PROBABILITY

In the special case p =

we can interpret these formulas in a particularly

simple way. Given a pattern A of m heads and tails, let

A:A =

fIkpl

[Ack’

=A(kj] .

k=l

(8.76)

We can easily find the binary representation of this number by placing a ‘1’

under each position such that the string matches itself perfectly when it is

superimposed on a copy of itself that has been shifted to start in this position:

= HTHTHHTHTH

A:A=(1000010101)2=-512+16+4+l

=533

HTHTHHTHTH J

HTHTHHTHTH

HTHTHHTH'TH J

HTHTHHTHTH

HTHTHHTHTH J

HTHTHHTHTH

HTHTHHTHTH J

Equation (8.74) now tells us that the expected number of flips until pattern A

appears is exactly 2(A:A), if we use a fair coin, because &kj = Ik when

p=q=$.

This result, first discovered by the Soviet mathematician A. D.

“Chem

bol’she

Solov’ev in 1966

[271],

seems paradoxical at first glance: Patterns with no

periodov

nasheg0

self-overlaps occur sooner

th,an

overlapping patterns do! It takes almost twice

s/ova,

tern

pozzhe

on0

poMl~ets%”

as long to encounter

HHHHH

as it does to encounter

HHHHT

THHHH.

-A.

Solov’ev

Now let’s consider an amusing game that was invented by (of all people)

Walter Penney

[231]

196!3.

Alice and Bill flip a coin until either HHT or

HTT occurs; Alice wins if the pattern HHT comes first, Bill wins if HTT comes

first. This game-now called “Penney ante” -certainly seems to be fair, if

played with a fair coin, because both patterns HHT and HTT have the same

characteristics if we look at them in isolation: The probability generating

function for the waiting tim’e until HHT first occurs is

G(z) =

8(2-

’

not! Who

and the same is true for HTT. Therefore neither Alice nor Bill has an

advan-

could they have an

tage, if they play solitaire.

advantage over?

8.4 FLIPPING COINS 395

But there’s an interesting interplay between the patterns when both are

considered simultaneously. Let

be the sum of Alice’s winning configura-

tions, and let Ss be the sum of Bill’s:

HHT

HHHT +

THHT

HHHHT + HTHHT + THHHT

. . .

;

Ss =

HTT

THTT

HTHTT

TTHTT

THTHTT

TTTHTT

+ . . . .

Also- taking our cue from the trick that worked when only one pattern was

involved-let us denote by N the sum of all sequences in which neither player

has won so far:

N = 1

+H+T+HH+HT+TH+TT+HHH+HTH+THH+...

(8.77)

Then we can easily verify the following set of equations:

l+N(H+T)

NfS~f.5.s;

NHHT

;

(8.78)

NHTT

= SATTS~.

If we now set H =

the resulting value of

becomes the probability

that Alice wins, and Ss becomes the probability that Bill wins. The three

equations reduce to

+N = N +sA +Ss;

s,;

+sg;

and we find

, Ss =

. Alice will win about twice as often as Bill!

In a generalization of this game, Alice and Bill choose patterns A and B

of heads and tails, and they flip coins until either A or B appears. The

two patterns need not have the same length, but we assume that A doesn’t

occur within B, nor does B occur within A. (Otherwise the game would be

degenerate. For example, if A =

and B =

THTH,

poor Bill could never win;

and if A = HTH and B =

TH,

both players might claim victory simultaneously.)

Then we can write three equations analogous to (8.73) and (8.78):

+N(H+T)

N+SA+S~;

NA =

A(lPkj [A

min(l,m)

(k’

=A(kj] +

sp,

A(lmk)

[Bck)

=Aiki];

k=l k=l

min(l,m)

NB =

lrnpk’

[Atk’

B(k)]

+ Ss

BCmPk)

[Bck)

= B,,,] .

k=l

(8.79)

396 DISCRETE PROBABILITY

Here

is the length of A and m is the length of B. For example, if we have

A = HTTHTHTH and B = THTHTTH, the two pattern-dependent equations are

N HTTHTHTH = SA TTHTHTH + SA

TTHTHTH +

THTH

;

N THTHTTH = SA THTTH + SA TTH +

THTTH +

We obtain the victory probabilities by setting H = T =

if we assume that a

fair coin is being used; this reduces the two crucial equations to

S/I

]Alk’

A.(k)]

+ Ss

[Bckl

Ackj]

;

k=l

(8.80)

N =SA

[Alk)

= B,,,] + Ss

[Bckl

B(k)]

k=l k=l

We can see what’s going on if we generalize the A:A operation of (8.76) to a

function of two independent strings A and B:

min(l,m)

A:B =

2kp’

[Alk’

=Bck,] .

k=l

Equations (8.80) now become simply

S*(A:A) + Ss(B:A) = S*(A:B) + Ss(B:B)

;

the odds in Alice’s favor are

B:B

B:A

A:A-A:B

(8.81)

(8.82)

(This beautiful formula was discovered by John Horton Conway

[ill].)

For example, if A = HTTHTHTH and B = THTHTTH as above, we have

A:A = (10000001)2 = 129, A:B = (0001010)2 = 10, B:A = (0001001)2 = 9,

and B:B = (1000010)2 = 66; so the ratio

SA/SB

(66-9)/(129-10)

= 57/l 19.

Alice will win this one only 57 times out of every 176, on the average.

Strange things can happen in Penney’s game. For example, the pattern

HHTH wins over the pattern HTHH with

3/2

odds, and HTHH wins over THHH with

7/5 odds. So HHTH ought to ‘be much better than THHH. Yet THHH actually wins

over HHTH, with 7/5 odds!

‘The

relation between patterns is not transitive. In

Odd, odd.

fact, exercise 57 proves that if Alice chooses any pattern

. .

of length

3 3, Bill can always ensure better than even chances of winning if he chooses

the pattern ;S2rlr2 . . .

~1~1,

where

is the heads/tails opposite of

~2.

8.5 HASHING 397

8.5 HASHING

Somehow the verb

“to

hash”

magically

became standard

terminology for key

transformation dor-

ing the mid-l

96Os,

yet nobody was rash

enough to use such

an undignified word

publicly until 1967.

-D.

Knoth

75)

Let’s conclude this chapter by applying probability theory to com-

puter programming. Several important algorithms for storing and retrieving

information inside a computer are based on a technique called “hashing!’

The general problem is to maintain a set of records that each contain a “key”

value, K, and some data D(K) about that key; we want to be able to find

D(K) quickly when K is given. For example, each key might be the name of

a student, and the associated data might be that student’s homework grades.

In practice, computers don’t have enough capacity to set aside one mem-

ory cell for every possible key; billions of keys are possible, but comparatively

few keys are actually present in any one application. One solution to the

problem is to maintain two tables KEY [jl and

DATACjl

for 1 6 j 6 N, where

N is the total number of records that can be accommodated; another vari-

able n tells how many records are actually present. Then we can search for a

given key K by going through the table sequentially in an obvious way:

Sl Set j :=

(We’ve searched through all positions < j.)

S2 If j > n, stop. (The search was unsuccessful.)

S3 If KEY Cjl = K, stop. (The search was successful.)

S4 Increase j by 1 and return to step S2. (We’ll try again.)

After a successful search, the desired data entry D(K) appears in

DATACjl.

After an unsuccessful search, we can insert K and D(K) into the table by

setting

j, KEY

Cnl

:= K,

DATACnl

:= D(K),

assuming that the table was not already filled to capacity.

This method works, but it can be dreadfully slow; we need to repeat

step S2 a total of n + 1 times whenever an unsuccessful search is made, and

n can be quite large.

Hashing was invented to speed things up. The basic idea, in one of its

popular forms, is to use m separate lists instead of one giant list. A “hash

function” transforms every possible key K into a list number h(K) between 1

and m. An auxiliary table

FIRSTCil

for 1 6 i 6 m points to the first record

in list

another auxiliary table

NEXTCjl

for 1 < j 6 N points to the record

following record j in its list. We assume that

FIRSTCi] = -1,

if list i is empty;

NEXT[jl = 0,

if record j is the last in its list.

As before, there’s a variable n that tells how many records have been stored

altogether.

398 DISCRETE PROBABILITY

For example, suppose

,the

keys are names, and suppose that there are

m = 4 lists based on the first letter of a name:

for ,4-F;

h(name) =

2, for G-L;

3, for M-R;

4, for

!3-Z.

We start with four empty lists and with n = 0. If, say, the first record has

Nora as its key, we have h(Nora) = 3, so Nora becomes the key of the first

item in list 3. If the next two names are Glenn and Jim, they both go into

list 2. Now the tables in memory look like this:

FIRST[l] =

-1,

FIRST[2] = 2, FIRST

[31

FIRST

[41

-1

KEY

Cl1

= Nora,

NEXT[l1

= 0;

KEY

[21

= Glenn,

NEXTC21

= 3;

KEY

[31

= Jim,

NEXTC31

= 0; n = 3.

(The values of DATA

[ll

DATA[21,

and

DATAC31

are confidential and will not

be shown.) After 18 records have been inserted, the lists might contain the

Let’s hear it for

names

the Concrete Math

students who sat in

list 1 list 2 list 3 list 4

the front rows and

lent their names to

Dianne

Ari

Brian

Fran

Doug

Glenn

Jim

Jennifer

Joan

Jerry

Jean

Nora

Mike

Michael

Ray

Paula

Scott

Tina

this experiment.

and these names would appear intermixed in the KEY array with NEXT entries

to keep the lists effectively separate. If we now want to search for John, we

have to scan through the six names in list 2 (which happens to be the longest

list); but that’s not nearly as bad as looking at all 18 names.

Here’s a precise specification of the algorithm that searches for key K in

accordance with this scheme:

Set i := h(K) and j := FIRSTCil.

H2 If j 6 0, stop. (The search was unsuccessful.)

H3 If KEY Cjl = K, stop. (The search was successful.)

H4 Set i := j, then set j

NEXTCi]

and return to step H2. (We’ll try again.)

For example, to search for Jennifer in the example given, step Hl would set

2 and j

2; step H3

,would

find that Glenn # Jennifer; step H4 would

bet their parents

set j := 3; and step H3 would find Jim # Jennifer.

are glad about that.

8.5 HASHING 399

After a successful search, the desired data D(K) appears in

DATA

[jl

, as in

the previous algorithm. After an unsuccessful search, we can enter K and D(K)

in the table by doing the following operations:

n :=

n+l;

if j < 0 then

FIRSTCil

:=n else

NEXT[il

:=n;

KEYCn.1

DATACnl

:= D(K);

NEXT[n]

:= 0.

(8.83)

Now the table will once again be up to date.

We hope to get lists of roughly equal length, because this will make the

task of searching about m times faster. The value of m is usually much greater

than 4, so a factor of l/m will be a significant improvement.

We don’t know in advance what keys will be present, but it is generally

possible to choose the hash function h so that we can consider h(K) to be a

random variable that is uniformly distributed between 1 and m, independent

of the hash values of other keys that are present. In such cases computing the

hash function is like rolling a die that has m faces. There’s a chance that all

the records will fall into the same list, just as there’s a chance that a die will

always turn up q ; but probability theory tells us that the lists will almost

always be pretty evenly balanced.

Analysis of Hashing: Introduction.

“Algorithmic analysis” is a branch of computer science that derives quan-

titative information about the efficiency of computer methods. “Probabilistic

analysis of an algorithm” is the study of an algorithm’s running time, con-

sidered as a random variable that depends on assumed characteristics of the

input data. Hashing is an especially good candidate for probabilistic analysis,

because it is an extremely efficient method on the average, even though its

worst case is too horrible to contemplate. (The worst case occurs when all

keys have the same hash value.) Indeed, a computer programmer who uses

hashing had better be a believer in probability theory.

Let P be the number of times step H3 is performed when the algorithm

above is used to carry out a search. (Each execution of H3 is called a “probe”

in the table.) If we know P, we know how often each step is performed,

depending on whether the search is successful or unsuccessful:

Step Unsuccessful search

Hl 1 time

H2 P + 1 times

H3 P times

H4 P times

Successful search

1 time

P times

1 times

400 DISCRETE PROBABILITY

Thus the main quantity that governs the running time of the search procedure

is the number of probes, P.

We can get a good mental picture of the algorithm by imagining that we

are keeping an address book that is organized in a special way, with room for

only one entry per page. On the cover of the book we note down the page

number for the first entry in each of m lists; each name K determines the list

h(K) that it belongs to.

Every

page inside the book refers to the successor

page in its list. The number of probes needed to find an address in such a

book is the number of pages we must consult.

If n items have been inserted, their positions in the table depend only

on their respective hash

val.ues,

(h’ , hz, . . . , &). Each of the m” possible

sequences (h’ , h2, . . . ,

is considered to be equally likely, and P is a random

variable depending on such a sequence.

Case 1: The key is not present.

Check under the

Let’s consider first the behavior of P in an unsuccessful search, assuming

doormat.

that n records have previously been inserted into the hash table. In this case

the relevant probability

spac:e

consists of

mn+’

elementary events

= (h’,hz,...,h,;hT,+‘)

where

is the hash value of the jth key inserted, and where &+’ is the hash

value of the key for which the search is unsuccessful. We assume that the

hash function h has been chosen properly so that

Pr(w)

= 1 /mnf’ for every

such

CU.

For example, if m = n

2, there are eight equally likely possibilities:

h3:

1:2

11 2:o

1 2

1:l

2 2: 1

2 1

1:l

2 1

2:l

2 2

1:o

2 2

212

If h’ =

we make two unsuccessful probes before concluding that the

new key K is not present; if h’ =

we make none; and so on. This list

of all possibilities shows that P has a probability distribution given by the pgf

(f +

$2’)

= (i + iz)‘, when m = n = 2.

An unsuccessful search makes one probe for every item in list number

n+‘,

so we have the general formula

P =

[h,

=hm+,l + [hz=hn+,l + ... + [h,,=hn+ll.

(8.84)

8.5 HASHING 401

The probability that

= hn+l is 1

/m,

for 1 < j 6 n; so it follows that

EP =

E[hl=~+l]+E[h~=hh,+,]+...tE[h,=h,+,]

Maybe we should do that more slowly: Let

be the random variable

ThenP=X1+...+X,,andEXj=l/mforallj<n;hence

EP =

EXl+...+EX,

n/m.

Good: As we had hoped, the average number of probes is

l/m

times what it

was without hashing. Furthermore the random variables

are independent,

and they each have the same probability generating function

Xj(Z) =

m-l+2

therefore the pgf for the total number of probes in an unsuccessful search is

P(z) =

(2). .

.X,(z)

(m-;+z)“.

This is a binomial distribution, with p

l/m and q = (m

1)/m;

in other

words, the number of probes in an unsuccessful search behaves just like the

number of heads when we toss a biased coin whose probability of heads is

l/m on each toss. Equation (8.61) tells us that the variance of P is therefore

n(m- 1)

npq =

mz *

When m is large, the variance of P is approximately n/m, so the standard

deviation is approximately

fi.

Case 2: The key is present.

Now let’s look at successful searches. In this case the appropriate proba-

bility space is a bit more complicated, depending on our application: We will

let

be the set of all elementary events

w = (h

,,...,

h,;k),

(8.86)

where

is the hash value for the jth key as before, and where k is the index

of the key being sought (the key whose hash value is hk). Thus we have

1 6

< m for 1 6 j < n, and 1 < k 6 n; there are

rn”.

n elementary

events w in all.

402 DISCRETE PROBABILITY

Let

be the probability that we are searching for the jth key that was

inserted into the table. Then

Pr(w) =

sk/mn

(8.87)

if w is the event (8.86). (Some applications search most often for the items

that were inserted first, or for the items that were inserted last, so we will not

assume that each

= l/n.) Notice that

,&l

Pr(w) =

Et=,

hence

(8.87) defines a legal probability distribution.

The number of probes P in a successful search is p if key K was the pth

key to be inserted into its hst. Therefore

P = [h, =

h-k]

+ [hz = hkl + . . . + [hk

=hkl

;

or, if we let

be the random variable [hj = hk], we have

= x1

“‘+xk.

(8.88)

Suppose, for example, that we have m = 10 and n = 16, and that the hash

values have the following “random” pattern:

Where

have I

seen

that pattern before?

(h-l,...,

h,6)=3

141592653589793;

(Pl,.

P,~)=1112111122312133.

The number of probes

needed to find the jth key is shown below hi.

Equation (8.88) represents P as a sum of random variables, but we can’t

simply calculate EP as EX,

$-.

. .+EXk because the quantity k itself is a random

variable. What is the probability generating function for P? To answer this

question we should digress

moment to talk about conditional probability.

Equation

(8.43) was

If A and B are events in a probability space, we say that the conditional

a1so

momentary

probability of A, given B, is

digression.

F’r(cu

A n B)

Pr(wEAIwEB)

Pr(wCB)

’

For example, if X and Y are random variables, the conditional probability of

the event X = x, given that Y = y, is

Pr(X=x and

Y=y)

Pr(X=xlY=y)

Pr(Y=y)

’

(8.90)

For any fixed y in the range of Y, the sum of these conditional probabil-

ities over all x in the range of X is Pr(Y

=y)/Pr(Y

=y) = 1; therefore (8.90)

defines a probability distribution, and we can define a new random variable

‘X/y’ such that Pr(Xly

=x)

= Pr(X =x 1 Y =y).

8.5 HASHING

303

If X and Y are independent, the random variable Xly will be essentially

the same as X, regardless of the value of y, because Pr(X = x 1 Y = y ) is equal

to Pr(X

=x)

by (8.5); that’s what independence means. But if X and Y are

dependent, the random variables X/y and Xly’ need not resemble each other

inanywaywheny#y’.

If X takes only nonnegative integer values, we can decompose its pgf into

a sum of conditional pgf’s with respect to any other random variable Y:

Gx(z)

WY=y)Gx,(z).

YEYIf~l

(8.91)

This holds because the coefficient of

on the left side is Pr(X

=x),

for all

X(n), and on the right it is

Pr(Y=y)Pr(x=xIY=y)

Pr(X=x and

Y=y)

YEytni YEYin)

= Pr(X=x).

For example, if X is the product of the spots on two fair dice and if Y is the

sum of the spots, the pgf for

X16

Gx,6(z) = +z5 + $z8 + ;z9

because the conditional probabilities for Y = 6 consist of five equally probable

events

{

m}.

Equation (8.91) in this case

reduces to

Gx(z)

2(z)

$x,3(z)

&Gx

z,(z)

$x,5(z)

j$x,dz)

$&T(Z)

j$x,a(z)

&Gx~9(4

$%Io(z)

j$x,,,

(~1

&12(z),

Oh, now

un-

de&and what

mathematicians

mean when they

say something is

“‘obvious,” “clear,”

or “trivial.”

a formula that is obvious once you understand it. (End of digression.)

In the case of hashing, (8.91) tells us how to write down the pgf for probes

in a successful search, if we let X = P and Y = K. For any fixed k between 1

and n, the random variable

k is defined as a sum of independent random

variables

+ . . . +

Xk;

this is (8.88).

it has the pgf

Gp,k(Z) =

(m-;+Z)k-‘Z.

404 DISCRETE PROBABILITY

Therefore the pgf for P itself is

GP(z)

‘&~GP,I;(z’I

k=l

(

n-l+2

’

where

(8.92)

S(z)

s2z

S,P’

(8.93)

is the pgf for the search probabilities

(divided by

for convenience).

Good. We have a probability generating function for P; we can now find

the mean and variance by differentiation. It’s somewhat easier to remove the

factor first, as we’ve done before, thus finding the mean and variance of

1 instead:

F(z)

Gp(z),‘z

S(m-m+f)

;

F’(z)

;S’(+)

;

F”(z) =

-&“(!+)

Therefore

EP = 1 + Mean(F) = 1 + F’( 1) = 1 + m-’ Mean(S) ;

(8.94)

VP =

Var(F)

F"(l)+F'(l)-F'(l)'

= rn-‘S”(1) +m-‘S’(1) -m~2S’(1)2

rnp2

Va.r(S)

(rn-’

m-*) Mean(S).

(8.95)

These are general

formula,s

expressing the mean and variance of the num-

ber of probes P in terms

‘of

the mean and variance of the assumed search

distribution S.

For example, suppose we have

= l/n for 1 6 k 6 n. This means

we are doing a purely

“ran.dom”

successful search, with all keys in the table

equally likely. Then

S(z)

is the uniform probability distribution U,(z) in

8.5 HASHING 405

(8.32),

and we have Mean(S) = (n-

1)/2,

Var(S)

(n2

1)/12.

Hence

n2-1

(m-l)(n-1)

(n-1)(6m+n-5)

VP==+

-Jm2

12m2

’

(8&v)

Once again we have gained the desired speedup factor of 1 /m. If m =

n/inn

and n

00,

the average number of probes per successful search in this case

is about

Inn, and the standard deviation is asymptotically

(Inn)/&!.

On the other hand, we might suppose that

= (kH,))’ for 1 6 k 6 n;

this distribution is called “Zipf’s law!’ Then Mean(G) = n/H,, and Var( G) =

in(n

+ 1)/H,,

n’/Hi.

The average number of probes for m =

n/inn

is approximately 2, with standard deviation asymptotic to

G/d.

In both cases the analysis allows the cautious souls among us, who fear

the worst case, to rest easily: Chebyshev’s inequality tells us that the lists

will be nice and short, except in extremely rare cases.

Case 2, continued: Variants of the variance.

We have just computed the variance of the number of probes in a success-

ful search, by considering P to be a random variable over a probability space

with

mn.n

elements

(h,,

. . .

hn;

k). But we could have adopted another point

OK, gang, time

of view: Each pattern

(h,

, . . . , h,) of hash values defines a random variable

to put

on your

skim suits again.

P/h,...

, h,), representing the probes we make in a successful search of a

-Friendly TA

particular hash table on n given keys. The average value of

(h, , . . . , h,),

A(h,, . . .

,&I

~p.Pr(Pl(hl,...,h,)=p),

(8.98)

p=l

can be said to represent the running time of a successful search. This quantity

A(h,, . . . , h,) is a random variable that depends only on (h, , . . . , h,), not on

the final component k; we can write it in the form

A(h,,...

,hn) =

$kPb,,...,hn;k),

k=l

since

P/(hl,...

, h,) = p with probability

~~=,

Pr(P(hl,...

,h,;k)=p)

= xE=, m nsk[P(hl,...

,h,;k)=p]

~~=,

Prh

, . . . , hn; k)

~~=,

nSk

fsk[P(h

I,...,

h,;k)=p].

k=l

406 DISCRETE PROBABILITY

The mean value of A(hl , . . . ,

&),

obtained by summing over all

m”

pos-

sibilities (hl , . . . ,

and dividing by mn, will be the same as the mean value

we obtained before in (8.g4), But the variance of A(hl , ,

h,)

is something

different; this is a variance of

averages, not a variance of

m”

probe

counts. For example, if m

1 (so that there is only one list), the “average”

value A(hl, . . .

,&)

A(l).

. . , 1) is actually constant, so its variance VA is

zero; but the number of probes in a successful search is not constant, so the

variance VP is nonzero.

But the VP is

We can illustrate this difference between variances by carrying out the

nonzero

“‘yin

calculations for general m and n in the simplest case, when

= l/n for

election year.

1 < k 6 n. In other words, we will assume temporarily that there is a uniform

distribution of search keys. Any given sequence of hash values (h, , . ,

defines m lists that contain respectively (n,

,nz,

. . .

,n,)

entries for some

numbers ni, where

nl+n2+...+n,

= n.

A successful search in which each of the n keys in the table is equally likely

will have an average running time of

(l+...+nl)

(l+...+nz)

+...+

(l+...+n,)

A(h,,...,h,)

(nlfl)

+ nz(n2+1) + . . +

n,(n,+l)

n:+n:+...+&+n

zz-

probes. Our goal is to calculate the variance of this quantity A(hl , . . . ,

&),

over the probability space

cionsisting

of all

m”

sequences

(hl

, . . . ,

h,).

The calculations will be simpler, it turns out, if we compute the variance

of a slightly different quantity,

B(h,,...,h,)

(?‘)+(T)+...+(y).

We have

A(h,,

. . .

+B(h,...,h,)/n,

hence the mean and

varianc:e

of A satisfy

EA =

1,;;

VA =

(8.99)

8.5 HASHING 407

The probability that the list sizes will be nl ,

n2,

. . . ,

is the multinomial

coefficient

(

nl,nz,...,n,

nl!n2! .

..n.!

divided by mn; hence the pgf for B( hl , . . . , h,) is

B,(z)

(

J;‘)+(;‘)+-+(“jq m-n.

,n2

,....n,>o

nl,nz,...,n,

+n2

t...+n,=n

This sum looks a bit scary to inexperienced eyes, but our experiences in

Chapter 7 have taught us to recognize it as an m-fold convolution. Indeed, if

we consider the exponential super-generating function

G(w,z)

~Bn,z,~,

n20

we can readily verify that G (w,

is simply an mth power:

As a check, we can try setting

= 1; we get G(w, 1) = (ew)m, so the coefficient

of m”w”/n! is

(1)

= 1.

If we knew the values of

B,/,

(1) and

(1))

we would be able to calculate

Var(B,). So we take partial derivatives of

G(w,

with respect to

&G(w,z) =

LB:,(z)?

7x30

m(&z(:)

%)me’

(i)Z(‘)Pl

w,z) =

xB;(z)y

3x30

408 DISCRETE PROBABILITY

Complicated, yes; but everything simplifies greatly when we set

For

example, we have

B;(l)y

m,+m

‘)y

n30

k>2

2(k

2)!

= me{”

‘)w

wkf2

k>O

2k!

,,z,im

llw

E--e

(

mw)n+2

n(n-l)m”wn

2mn! =

TX30

II30

2mn!

’

and it follows that

The expression for EA in

(8.!3g)

now gives EA = 1 + (n-

1)/2m,

in agreement

with (8.96).

The formula for Bz (1) involves the similar sum

(;) (G>-,)

(k+

‘)k(k;;)(-“

hence we find that

= mewm(+mw4 + w”)

;

B;(l)

(;:)((1)

-l>s*

(8.101)

Now we can put all the pieces together and evaluate the desired variance VA.

Massive cancellation occurs, and the result is surprisingly simple:

B”(1) +

B;(l)

B;(1)2

VA+L

=--

(n+l)(n-2)

+-m-n(n-1)

2 4

(m-

l)i:n-

2mln

(8.102)

8.5 HASHING 409

When such “coincidences” occur, we suspect that there’s a mathematical

reason; there might be another way to attack the problem, explaining why

the answer has such a simple form. And indeed, there is another approach (in

exercise 60), which shows that the variance of the average successful search

has the general form

VA =

k=l

(8.103)

when

is the probability that the kth-inserted element is being sought.

Equation (8.102) is the special case

= l/n for 1 < k 6 n.

Besides the variance of the average, we might also consider the average of

the variance. In other words, each sequence (hl , . . . ,

hn)

that defines a hash

table also defines a probability distribution for successful searching, and the

variance of this probability distribution tells how spread out the number of

probes will be in different successful searches. For example, let’s go back to

Where have

seen

the case where we inserted n = 16 things into m = 10 lists:

that pattern before?

Where have

seen

, . . .

,h16)=3

141592653589793

that grafito before?

IqvP,

Pl,

,P,6)=1112111122312133

A successful search in the resulting hash table has the pgf

G(3,1,4,1,...

,3)

SkZP(3,1,4,1,...,3;k)

k=l

SlZ+S2Z+S3Z+S4Z2+...+S~~Z3.

We have just considered the average number of probes in a successful search

of this table, namely

A(3,1,4,1,.

. . ,3) =

Mean(G(3,1,4,1,.

. .

,3)).

We can

also consider the variance,

This variance is a random variable, depending on (hl , . . . , h,), so it is natural

to consider its average value.

In other words, there are three natural kinds of variance that we may

wish to know, in order to understand the behavior of a successful search: The

overuZZ

variance of the number of probes, taken over all (h1,. , . , h,,) and k;

the variance of the average number of probes, where the average is taken

over all k and the variance is then taken over all (h, , . . . , h,,); and the average

of the variance of the number of the probes, where the variance is taken over

410 DISCRETE PROBABILITY

all k and the average is

the:n

taken over all (hi,. , h,). In symbols, the

overall variance is

Vf’

-+J(h,,...,h,;k)’

l<h,

,...,

h,,$m

k=l

f-$Ph.....hn;k))2;

l$h

I,...,

h,,<m

k=l

the variance of the average is

and the average of the variance is

AV =

SkP(h,,...,h,;k)2

lsh

,I..,,

h,,$m

SkP(h,,...,h,;k)

It turns out that these three quantities are interrelated in a simple way:

VP =

VA+AV.

(8.104)

In fact, conditional probability distributions always satisfy the identity

= V(E(XlY)) + E(V(XlY))

(8.105)

if X and Y are random variables in any probability space and if X takes real

values.

(This identity is proved in exercise 22.) Equation (8.104) is the

special case where X is the number of probes in a successful search and Y is

the sequence of hash values (hl , . . . , h,).

The general equation (8.105) needs to be understood carefully, because

the notation tends to conceal the different random variables and probability

spaces in which expectations and variances are being calculated. For each y

in the range of Y, we have defined the random variable Xly in

(~.Qo),

and this

random variable has an expected value E(Xly) depending on

Now E(XlY)

denotes the random variable whose values are E(

)

as y ranges over all

8.5 HASHING 411

possible values of Y, and V(E(XlY)) is th

variance of this random variable

[Now is a good

with respect to the probability distribution of Y. Similarly, E(V(XlY)) is the

time to do warmup

exercise 6.)

average of the random variables V(Xly) as y varies. On the left of (8.105)

VX,

the unconditional variance of X. Since variances are nonnegative, we

always have

vx 3

V(EW’))

and VX 3

E(V(XlY)).

(8.106)

Case 1, again: Unsuccessful search revisited.

Let’s bring our microscopic examination of hashing to a close by doing one

more calculation typical of algorithmic analysis. This time we’ll look more

closely at the total running time associated with an unsuccessful search,

assuming that the computer will insert the previously unknown key into its

memory.

P is still the num-

ber of probes.

The insertion process in (8.83) has two cases, depending on whether j is

negative or zero. We have j < 0 if and only if P = 0, since a negative value

comes from the FIRST entry of an empty list. Thus, if the list was previously

empty, we have P = 0 and we must set FIRSTC&+,l := n + 1. (The new

record will be inserted into position n + 1.) Otherwise we have P > 0 and we

must set a LINK entry to n +

These two cases may take different amounts

of time; therefore the total running time for an unsuccessful search has the

form

T = a+pP$-6[P=O],

(8.107)

where

OL,

fi,

and 6 are constants that depend on the computer being used and

on the way in which hashing is encoded in that machine’s internal language.

It would be nice to know the mean and variance of T, since such information

is more relevant in practice than the mean and variance of P.

So far we have used probability generating functions only in connection

with random variables that take nonnegative integer values. But it turns out

that we can deal in essentially the same way with

Gx(z) =

Pr(w)zx(wi

wcn

when X is any real-valued random variable, because the essential characteris-

tics of X depend only on the behavior of Gx near z =

where powers of z are

well defined. For example, the running time (8.107) of an unsuccessful search

is a random variable, defined on the probability space of equally likely hash

values (h1,. . , ,

h,;

h,+l

) with 1 6

6 m; we can consider the series

GT(z)

f...f

Z”+PPlhl

,...,

hn;hn+l)+6P(hl

a...>

hn;hn+l

I=01

h, =l

h,=l

h,+,=l

412 DISCRETE PROBABILITY

to be a pgf even when

01,

(3,

and 6 are not integers. (In fact, the parameters

(3,

6 are physical quantitieis that have dimensions of time; they aren’t even

pure numbers! Yet we can use them in the exponent of 2.) We can still

calculate the mean and variance of T, by evaluating G;( 1) and Gf’( 1) and

combining these values in the usual way.

The generating function for P instead of T is

P(z)

(

m-l+z)n

q =

xPr(P=p)zP

P>O

Therefore we have

a((~6-1)Pr’~P=O)+~Pr(P=p)zBP)

P20

The determination of

Mean

and

Var(G’)

is now routine:

Mean

= Gf(1) =

a+pt

+6(y)n;

Gt’(l) =

a(a-l)i-2ap~+B(B-l)~+lJ

2n(n-

+2a6(~)“+b(h-l)(q)“;

-G;(l)’

V=(G)

= Gf’(l) +

Gf-(

2n(m-

-24gn;

(8.108)

+b2((v)“--

(%)‘“).

(8.109)

In Chapter 9 we will

le’arn

how to estimate quantities like this when

m and n are large. If, for example, m = n and n

00, the techniques

of Chapter 9 will show that the mean and variance of T are respectively

oL+@+6e~‘+O(n~‘) and

~2--2@6ee’+62(e~‘-e~2)+O(n~‘).

Ifm =

n/inn

and n -+

the corresponding results are

Mean

(31nn+a+6/n+O((logn)2/n2);

Var(G’) = (S21nn-

((/31nn)2+2~61nn-62)/n+O((logn)3/n2),

8 EXERCISES 413

Exercises

Warmups

What’s the probability of doubles in the probability distribution

Pro,

(8.3),

when one die is fair and the other is loaded? What’s the proba-

bility that S = 7 is rolled?

What’s the probability that the top and bottom cards of a randomly shuf-

fled deck are both aces? (All

52!

permutations have probability l/52!.)

Stanford’s Concrete Math students were asked in 1979 to flip coins until

Why only ten they got heads twice in succession, and to report the number of flips

numbers?

required. The answers were

The other students

either weren’t

3, 2, 3, 5,

‘IO,

2, 6, 6, 9, 2.

empiricists or

they were just too

Princeton’s

Co:ncrete

Math students were asked in 1987 to do a similar

Aipped out.

thing, with the following results:

10, 2, 10, 7, 5, 2, 10, 6, 10, 2.

Estimate the mean and variance, based on (a) the Stanford sample;

(b) the Princeton sample.

Let H(z) =

F(z)/G(z),

where F(1) = G(1) = 1. Prove that

Mean(H) = Mean(F) -Mean(G),

Var(H) =

Var(F)

Var(G)

in analogy with (8.38) and

(8.3g),

if the indicated derivatives exist at

Suppose Alice and Bill play the game (8.78) with a biased coin that comes

up heads with probability p. Is there a value of p for which the game

becomes fair?

What does the conditional variance law (8.105) reduce to, when X and Y

are independent random variables?

Basics

Show that if two dice are loaded with the same probability distribution,

the probability of doubles is always at least

Let A and B be events such that A U B =

f2.

Prove that

Pr(wEAClB) =

Pr(wEA)Pr(wEB)-Pr(w$A)Pr(w$B).

Prove or disprove: If X and Y are independent random variables, then so

are F(X) and G(Y), when F and G are any functions.

414 DISCRETE PROBABILITY

What’s the maximum number of elements that can be medians of a ran-

dom variable X, according to definition (8.7)?

Construct a random variable that has finite mean and infinite variance.

12 a If P(z) is the pgf for the random variable X, prove that

Pr(X $ r) < x.~‘P(x) for 0 <

< 1;

Pr(X 3 r) 6 x. -‘P(x)

for x 3 1.

(These important relations are called the tail inequalities.)

b In the special case P(z) = (1

+~)“/2~,

use the first tail inequality to

prove that

k,,,(z)

l/xan(l

CX)~'-~)~

when 0 <

OL<

IfX,,

..)

Xln are inde:pendent random variables with the same distri-

bution, and if

is any real number whatsoever, prove that

x1+...+xzn

X1+-'fX,-K

IL1

14 Let F(z) and G(z) be probability generating functions, and let

H(z) = pF(z) +

G(z)

where p + q =

(This is called a miztzlre of F and G; it corresponds to

flipping a coin and choosing probability distribution F or G depending on

whether the coin comes up heads or tails.) Find the mean and variance

of H in terms of

and the mean and variance of F and G.

15 If F(z) and G(z) are probability generating functions, we can define an-

other pgf H(z) by “composition”:

H(z) =

F(G(z)).

Express Mean(H) and Var(H) in terms of Mean(F), Var(F), Mean(G),

and Var(G). (Equation (8.92) is a special case.)

16 Find a closed form for the super generating function

En20

Fn(z)wn,

when F,(z) is the football-fixation generating function defined in (8.53).

17 Let

X,,,

and

Yn,p

have the binomial and negative binomial distributions,

respectively, with parameters (n, p). (These distributions are defined in

(8.57) and (8.60).) Prove that

Pr(Y,,,

<m)

= Pr(Xm+n,p

an).

What

identity in binomial

coe,fficients

does this imply?

18 A random variable X is said to have the Poisson distribution with

The distribution of

mean

if Pr(X= k) = eeppk/k! for all k 3 0.

fish

per unit volume

What is the pgf of such a random variable?

of water.

b What are its mean, variance, and other cumulants?

8 EXERCISES 415

19 Continuing the previous exercise, let

be a random Poisson variable

with mean

~1,

and let

be a random Poisson variable with mean

~2,

independent of

X1.

What is the probability that

= n?

What are

t.he

mean, variance, and other cumulants of 2x1 + 3X2?

20 Prove (8.74) and (8.75), the general formulas for mean and variance of

the time needed to wait for a given pattern of heads and tails.

21 What does the value of N represent, if H and T are both set equal to

in (8.77)?

22 Prove (8.105)~ the law of conditional expectations and variances.

Homework exercises

23 Let

Pro0

be the probability distribution of two fair dice, and let

Prll

the probability distribution of two loaded dice as given in (8.2). Find all

events A such that

Proo(A)

Prll

(A). Which of these events depend

only on the random variable S? (A probability space with

has

236

events; only 2” of those events depend on S alone.)

24 Player J rolls

2n+

1 fair dice and removes those that come up q . Player

K then calls a number between 1 and 6, rolls the remaining dice, and

removes those that show the number called. This process is repeated

until no dice remain. The player who has removed the most total dice

(n + 1 or more) is the winner.

What are the mean and variance of the total number of dice that

J removes? Hint: The dice are independent.

What’s the probability that J wins, when n = 2?

25 Consider a gambling game in which you stake a given amount A and you

roll a fair die. If k spots turn up, you multiply your stake by 2(k

1)/5.

(In particular, you double the stake whenever you roll q , but you lose

everything if you roll q .) You can stop at any time and reclaim the

current stake. What are the mean and variance of your stake after n rolls?

(Ignore any effects of rounding to integer amounts of currency.)

26 Find the mean and variance of the number of L-cycles in a random permu-

tation of n elements. (The football victory problem discussed in (8.23),

(8.24), and (8.53) is the special case

= 1.)

27 Let

X1,

X,7,

. . . , X, be independent samples of the random variable X.

Equations (8.19) and (8.20) explain how to estimate the mean and vari-

ance of X on the basis of these observations; give an analogous formula

for estimating the third cumulant ~3. (Your formula should be an “un-

biased” estimate, in the sense that its expected value should be KS.)

416 DISCRETE PROBABILITY

28 What is the average length of the coin-flipping game (8.78)

given that Alice wins?

b given that Bill wins?

29 Alice, Bill, and Computer flip a fair coin until one of the respective

patterns A = HHTH, B

HTHH, or C = THHH appears for the first time.

(If only two of these patterns were involved, we know from (8.82) that A

would probably beat B, that B would probably beat C, and that C would

probably beat A; but all three patterns are simultaneously in the game.)

What are each player’s chances of winning?

30 The text considers three kinds of variances associated with successful

search in a hash table. Actually there are two more: We can consider the

average (over k) of the variances (over

, . . . ,

h,)

of P(

, . . . ,

h,;

k); and

we can consider the variance (over k) of the averages (over

hl,

. . . , h,,).

Evaluate these quantities.

31 An apple is located at vertex A of pentagon ABCDE, and a worm is

located two vertices away, at C. Every day the worm crawls with equal

probability to one of the two adjacent vertices. Thus after one day the

worm is at vertex B with probability

and at vertex D with probability

Schrtidinger’s worm.

After two days, the worm might be back at C again, because it has no

memory of previous positions. When it reaches vertex A, it stops to dine.

What are the mean and variance of the number of days until dinner?

b Let p be the probability that the number of days is 100 or more.

What does Chebyshev’s inequality say about p?

What do the tail inequalities (exercise 12) tell us about p?

32 Alice and Bill are in

t:he

military, stationed in one of the five states

Kansas, Nebraska, Missouri, Oklahoma, or Colorado. Initially Alice is in

Nebraska and Bill is in Oklahoma. Every month each person is reassigned

to an adjacent state, each adjacent state being equally likely. (Here’s a

diagram of the adjacencies:

The initial states are circled.) For example, Alice is restationed after the

Definitely a finite-

first month to Colorado., Kansas, or Missouri, each with probability

l/3.

state

situation.

Find the mean and variance of the number of months it takes Alice and

Bill to find each other. (You may wish to enlist a computer’s help.)

(Use a calculator for

the numerical work

on this problem.)

8 EXERCISES 417

Are the random variables

and

in (8.88) independent?

Gina is a golfer who has probability p = .05 on each stroke of making a

“supershot” that gains a stroke over par, probability q =

.91

of making

an ordinary shot, and probability

= .04 of making a “subshot” that

costs her a stroke with respect to par. (Non-golfers: At each turn she

advances 2, 1, or 0 steps toward her goal, with probability

q, or r,

respectively. On a par-m hole, her score is the minimum n such that she

has advanced m or more steps after taking n turns. A low score is better

than a high score.)

Show that Gina wins a par-4 hole more often than she loses, when

she plays against a player who shoots par. (In other words, the

probability that her score is less than 4 is greater than the probability

that her score is greater than 4.)

Show that her average score on a par-4 hole is greater than 4. (There-

fore she tends to lose against a “steady” player on total points, al-

though she would tend to win in match play by holes.)

Exam

problems

35 A die has been loaded with the probability distribution

WFJ)

= PI

;

Pr(m)

= ~2; . . . .

Pr(m)

= p6.

Let

be the sum of the spots after this die has been rolled n times. Find

a necessary and sufficient condition on the “loading distribution” such

that the two random variables

mod 2 and

mod 3 are independent

of each other, for all n.

36 The six faces of a certain die contain the spot patterns

q pJUHpJH

instead of the usual

through

Show that there is a way to assign spots to the six faces of another

die so that, when these two dice are thrown, the sum of spots has the

same probability distribution as the sum of spots on two ordinary

dice. (Assume that all 36 face pairs are equally likely.)

Generalizing, find all ways to assign spots to the 6n faces of n dice so

that the distribution of spot sums will be the same as the distribution

of spot sums on n ordinary dice. (Each face should receive a positive

integer number of spots.)

37 Let p,, be the probability that exactly n tosses of a fair coin are needed

before heads are seen twice in a row, and let

,&,

pk. Find closed

forms for both p,, and

in terms of Fibonacci numbers.

418 DISCRETE PROBABILITY

What is the probability generating function for the number of times you

need to roll a fair die until all six faces have turned up? Generalize to

m-sided fair dice: Give closed forms for the mean and variance of the

number of rolls needed to see

of the m faces. What is the probability

that this number will be exactly n?

A Dirichlet probability generating function has the form

P(z) =

lI>l

Thus P(0) = 1. If X is a random variable with Pr(X=n) =

pn,

express

E(X), V(X), and E(lnX) in terms of P(z) and its derivatives.

The mth cumulant

of the binomial distribution (8.57) has the form

nfm(p), where

is a polynomial of degree m. (For example, fl (p) = p

and fz(p) = p

p2,

because the mean and variance are np and npq.)

Find a closed form for the coefficient of

in f,,,(p).

b Prove that

f,(i)

(2”

l)B,/m+

[m=ll,

where

is the mth

Bernoulli number.

Let the random variable X, be the number of flips of a fair coin until heads

have turned up a total of n times. Show that

E(X;:,)

(-l)n(ln2+

Hjnlz,

H,). Use the rnethods of Chapter 9 to estimate this value with

an absolute error of 0 (

?tp3

A certain man has a problem finding work. If he is unemployed on

any given morning, there’s constant probability

(independent of past

history) that he will be hired before that evening; but if he’s got a job

when the day begins, there’s constant probability pf that he’ll be laid

Does

7)$

choose

off by nightfall. Find the average number of evenings on which he will

optima’line

breaks?

have a job lined up, assuming that he is initially employed and that this

process goes on for n days. (For example, if n = 1 the answer is 1

-pi.)

Find a closed form for the pgf G,(z) =

tk3c

pk,nzk, where pk,n is the

probability that a random permutation of n objects has exactly k cycles.

What are the mean and standard deviation of the number of cycles?

The athletic department runs an intramural “knockout tournament” for

2” tennis players as follows. In the first round, the players are paired off

randomly, with each pairing equally likely, and

2nm

’

matches are played.

The winners advance to the second round, where the same process pro-

duces 2”

’

winners. And so on; the kth round has

2npk

randomly chosen

matches between the

2”-mkf’

players who are still undefeated. The nth

round produces the champion. Unbeknownst to the tournament organiz-

ers, there is actually an (ordering among the players, so that

is best, x2

A peculiar set of

tennis players.

‘A

fast arithmetic

computation shows

that the sherry is

always at least three

years old. Taking

computation further

gives the vertigo.”

-Revue du vin de

France

(Nov

1984)

8 EXERCISES 419

is second best, . . .

, x2” is worst. When

plays xk and j < k, the winner

with probability p and

with probability 1

independent of

the other matches. We assume that the same probability p applies to all

j and k.

What’s the probability that

wins the tournament?

b What’s the probability that the nth round (the final match) is be-

tween the top two players,

and x2?

C What’s the probability that the best

players are the competitors

in the kth-to-last round? (The previous questions were the cases

k=O and k= 1.)

d Let N(n) be the number of essentially different tournament results;

two tournaments are essentially the same if the matches take place

between the same players and have the same winners. Prove that

N(n) = 2”!.

What’s the probability that x2 wins the tournament?

Prove that if

< p < 1, the probability that xj wins is strictly

greater than the probability that xj+l wins, for 1 6 j < 2”.

45 True sherry is made in Spain according to a multistage system called

“Solera!’ For simplicity we’ll assume that the winemaker has only three

barrels, called A, B, and C. Every year a third of the wine from barrel C

is bottled and replaced by wine from B; then B is topped off with a third

of the wine from A; finally A is topped off with new wine. Let A(z), B(z),

C(z) be probability generating functions, where the coefficient of

Z”

the fraction of n-year-old wine in the corresponding barrel just after the

transfers have been made.

a Assume that the operation has been going on since time immemorial,

so that we have a steady state in which A(z), B(z), and C(z) are the

same at the beginning of each year. Find closed forms for these

generating functions.

b Find the mean and standard deviation of the age of the wine in each

barrel, under the same assumptions. What is the average age of the

sherry when it is bottled? How much of it is exactly 25 years old?

Now take the finiteness of time into account: Suppose that all three

barrels contained new wine at the beginning of year 0. What is the

average age of the sherry that is bottled at the beginning of year n?

46 Stefan Banach used to carry two boxes of matches, each containing

n matches initially. Whenever he needed a light he chose a box at ran-

dom, each with probability

independent of his previous choices. After

taking out a match he’d put the box back in its pocket (even if the box

became empty-all famous mathematicians used to do this). When his

chosen box was empty he’d throw it away and reach for the other box.

420 DISCRETE PROBABIL1T.Y

a Once he found that the other box was empty too. What’s the prob-

ability that this occurs? (For n = 1 it happens half the time and

for n = 2 it happens

3/B

of the time.) To answer this part, find a

closed form for the generating function P(w, z) =

t,,,

pm,nwmzn,

where pm,,,

is the probability that, starting with m matches in one

box and n in the other, both boxes are empty when an empty box

is first chosen. Then. find a closed form for P~,~.

b Generalizing your

a.nswer

to part (a), find a closed form for the

probability that exactly k matches are in the other box when an

empty one is first th.rown away.

Find a closed form for the average number of matches in that other

And for the number

box.

in the empty box.

4’7

Some physicians, collaborating with some physicists, recently discovered

a pair of microbes that reproduce in a peculiar way. The male microbe,

called a diphage, has two receptors on its surface; the female microbe,

called a triphage, has three:

diphage:

triphage:

receptor: 0

When a culture of diphages and triphages is irradiated with a psi-particle,

exactly one of the receptors on one of the phages absorbs the particle;

each receptor is equally likely. If it was a diphage receptor, that diphage

changes to a triphage; if it was a triphage receptor, that triphage splits

into two diphages. Thus if an experiment starts with one diphage, the

first psi-particle changes it to a triphage, the second particle splits the

triphage into two diphages, and the third particle changes one of the

diphages to a triphage. The fourth particle hits either the diphage or

the triphage; then there are either two triphages (probability

or three

diphages (probability

i).

Find a closed form for the average number

of diphages present, if we begin with a single diphage and irradiate the

culture n times with single psi-particles.

48 Five people stand at the vertices of a pentagon, throwing frisbees to each

Or, if this pentagon

other.

is in Arlington,

throwing missiles

at each other.

8 EXERCISES 421

Frisbee is a trade-

mark of Wham-O

Manufacturing

Company.

They have two frisbees, initially at adjacent vertices as shown. In each

time interval, each frisbee is thrown either to the left or to the right

(along an edge of the pentagon) with equal probability. This process

continues until one person is the target of two frisbees simultaneously;

then the game stops. (All throws are independent of past history.)

Find the mean and variance of the number of pairs of throws.

b Find a closed form for the probability that the game lasts more than

100 steps, in terms of Fibonacci numbers.

49 Luke Snowwalker spends winter vacations at his mountain cabin. The

front porch has m pairs of boots and the back porch has n pairs. Every

time he goes for a walk he flips a (fair) coin to decide whether to leave

from the front porch or the back, and he puts on a pair of boots at that

porch and heads off. There’s a 50/50 chance that he returns to each

porch, independent of his starting point, and he leaves the boots at the

porch he returns to. Thus after one walk there will be m + [-1 , 0, or

+l]

pairs on the front porch and n

[+l,

0, or -11 pairs on the back porch.

If all the boots pile up on one porch and if he decides to leave from

the other, he goes without boots and gets frostbite, ending his vacation.

Assuming that he continues his walks until the bitter end, let

(m,

n) be

the probability that he completes exactly N nonfrostbitten trips, starting

with m pairs on the front porch and n on the back. Thus, if both m

and n are positive,

PN(m,n) =

+PN-r(m-

l,n+l)

tPi+,(m,n)

this follows because this first trip is either front/back, front/front, back/

back, or back/front, each with probability

and N

1 trips remain.

a Complete the recurrence for

(m, n) by finding formulas that hold

when m = 0 or n = 0. Use the recurrence to obtain equations that

hold among the probability generating functions

gm,n(z) =

PN(myn)zN

N>O

b Differentiate your equations and set

= 1, thereby obtaining rela-

tions among the quantities

g&(

1). Solve these equations, thereby

determining the mean number of trips before frostbite.

C Show that

gm,n

has a closed form if we substitute

= 1

/cos2

gm,n

(

COG

sin(2m +

1)O

+ sin(2n + 1 )e

cos

sin(2m + 2n +

218

422 DISCRETE PROBABILITY

50 Consider the function

H(z) =

3+&-2)(9-z)).

The purpose of this problem is to prove that H(z) = tkZO

hkzk

is a

probability generating function, and to obtain some basic facts about it.

Let (1 -z)~/~(~-z)‘/~ =

k>O

ckzk.

Prove that

= 3,

= -14/3,

37/27,

and

c3+~

= 3 x,‘(k)

(l$)

($)

k+3

for all

3 0. Hint: Use

the identity

(9%z)"2 =

3(1

-2)"2(1

$z/(l

-2))"2

and expand the last factor in powers of z/( 1

z).

b Use part (a) and exercise 5.81 to show that the coefficients of H(z)

are all positive.

Prove the amazing identity

/=++2.

d What are the mean and variance of H?

51 The state lottery in El

Dorado

uses the payoff distribution H defined

in the previous problem. Each lottery ticket costs 1 doubloon, and the

payoff is k doubloons with probability

hk.

Your chance of winning with

each ticket is completely independent of your chance with other tickets;

in other words, winning or losing with one ticket does not affect your

probability of winning with any other ticket you might have purchased

in the same lottery.

Suppose you start with one doubloon and play this game. If you win

k doubloons, you buy k tickets in the second game; then you take

the total winnings in the second game and apply all of them to the

third; and so on.

1:f

none of your tickets is a winner, you’re broke

and you have to stop gambling. Prove that the pgf of your current

holdings after n rounds of such play is

‘-

dm14z)+2n-1

J(S-z)/(l

-z)+2n+l

’

b Let

be the probability that you lose all your money for the first

time on the nth game, and let

G(z)

glz

g2z2

+ ... . Prove

that G(1) = 1. (This means that you’re bound to lose sooner or

later, with probability

although you might have fun playing in

the meantime.) What are the mean and the variance of G?

8 EXERCISES 423

What is the average total number of tickets you buy, if you continue

to play until going broke?

d What is the average number of games until you lose everything if

A doubledoubloon.

you start with two doubloons instead of just one?

Bonus problems

Show that the text’s definitions of median and mode for random variables

correspond in some meaningful sense to the definitions of median and

mode for sequences, when the probability space is finite.

Prove or disprove: If X, Y, and Z are random variables with the property

that all three pairs (X, Y), (X, Z) and (Y, Z) are independent, then X + Y

is independent of Z.

Equation (8.20) proves that the average value of \iX is VX. What is the

variance of

VX?

A normal deck of playing cards contains 52 cards, four each with face

values in the set

{A,2,3,4,5,6,7,8,9,1O,J,Q,K}.

Let X and Y denote

the respective face values of the top and bottom cards, and consider the

following algorithm for shuffling:

Permute the deck randomly so that each arrangement occurs with

probability l/52!.

If X # Y, flip a biased coin that comes up heads with probability p,

and go back to step Sl if heads turns up. Otherwise stop.

Each coin flip and each permutation is assumed to be independent of all

the other randomizations. What value of p will make X and Y indepen-

dent random variables after this procedure stops?

Generalize the frisbee problem of exercise 48 from a pentagon to an

n-gon. What are the mean and variance of the number of collision-free

throws in general, when the frisbees are initially at adjacent vertices?

Show that, if m is odd, the pgf for the number of throws can be written

as a product of coin-flipping distributions:

(m-1

l/2

Pk=

G,(z) =

k=,

-qkz’

where

= sin

(2k-

1)~

cos2

‘2k2-; In.

Hint: Try the substitution z =

l/cos2

Prove that the Penney-ante pattern

‘~1~2

. . .

~~~1~1

is always inferior to

the pattern jszrlr2 . . .

‘cl-1

when a fair coin is flipped, if

3 3.

424 DISCRETE PROBABILITY

58 Are there patterns A and B of heads and tails such that A is longer

than B, yet A appears before B more than half the time when a fair coin

is being flipped?

Let k and n be fixed positive integers with k < n.

Find a closed form for the probability generating function

G(w,z) =

...

WP/hl

,....h,;klZP(hl

,....h,,;nl

h,=l

for the joint distribution of the numbers of probes needed to find the

kth and nth items that have been inserted into a hash table with

m lists.

Although the random variables P(h1,. . . ,

h,;

k) and P(h1,. . . , h,,; n)

are dependent, show that they are somewhat independent:

E(P(h,

, . . . ,

h,;

k)Ph,

. . . ,

n))

= (EP(h+...

h,;

k))

(Whl,

. . . , hn;

n))

60 Use the result of the previous exercise to prove (8.103).

Continuing exercise 47, find the variance of the number of diphages after

n irradiations.

Research

problems

62 The normal distribution is a non-discrete probability distribution char-

acterized by having all its cumulants zero except the mean and the vari-

ance. Is there an easy way to tell if a given sequence of cumulants

(Kl,Kz,K3,...)

comes from a discrete distribution? (All the probabil-

ities must be “atomic” in a discrete distribution.)

63 Is there any sequence A =

‘~1~2

. . . ri-1ri of

3 3 heads and tails such

that the sequences

Hr1

r.2

. . .

~1-1

and

Trlrr

. . .

‘cl-1

both perform equally

well against A in the game of Penney ante?

Asymptotics

EXACT ANSWERS are great when we can find them; there’s something

very satisfying about complete knowledge. But there’s also a time when

approximations are in order. If we run into a sum or a recurrence whose

solution doesn’t have a closed form (as far as we can tell), we still would like

to know something about the answer; we don’t have to insist on all or nothing.

And even if we do have a closed form, our knowledge might be imperfect, since

we might not know how to compare it with other closed forms.

For example, there is (apparently) no closed form for the sum

But it is nice to know that

oh . . here we say that the sum is “asymptotic to” 2(3,“). It’s even nicer to have more

comes that A-word.

detailed information, like

(3(2-;+0($)).

(9.1)

which gives us a “relative error of order 1 /n’.” But even this isn’t enough to

tell us how big

is, compared with other quantities. Which is larger,

the Fibonacci number

Fan?

Answer: We have

= 22 >

= 21 when n = 2;

but

Fan

is eventually larger, because

F4,,

$4n/&

and +4

6.8541, while

/36.751”(1

O($))

(9.2)

Our goal in this chapter is to learn how to understand and to derive results

like this without great pain.

425

426 ASYMPTOTICS

The word asymptotic stems from a Greek root meaning “not falling

Other words like

together!’ When ancient Greek mathematicians studied conic sections, they

‘symptom’

and

considered hyperbolas like the graph of y =

dm,

‘ptomaine’

also

come from this root.

which has the lines y =

and y = --x as “asymptotes!’ The curve approaches

but never quite touches these asymptotes, when x

00. Nowadays we use

“asymptotic” in a broader sense to mean any approximate value that gets

closer and closer to the truth, when some parameter approaches a limiting

value. For us, asymptotics means “almost falling together!’

Some asymptotic formulas are very difficult to derive, well beyond the

scope of this book. We will content ourselves with an introduction to the sub-

ject; we hope to acquire a suitable foundation on which further techniques can

be built. We will be particularly interested in understanding the definitions

‘m’

and ‘0’ and similar symbols, and we’ll study basic ways to manipulate

asymptotic quantities.

9.1 A HIERARCHY

Functions of n that occur in practice usually have different “asymp-

totic growth ratios”; one of them will approach infinity faster than another.

We formalize this by saying that

f(n)

g(n)

’

f(n)

&z&s(n)

(9.3)

This relation is transitive: If f(n) 4 g(n) and g(n) 4 h(n) then f(n) 4 h(n).

We also may write g(n) + f(n) if f(n) + g(n) . This notation was introduced

A/l

functions

in 1871 by Paul du Bois-Re:ymond

[29].

great and small.

For example, n 4

n’;

informally we say that n grows more slowly

than

n*.

In fact,

(9.4)

when a and

are arbitrary real numbers.

There are, of course, many functions of n besides powers of n. We can

use the + relation to rank lots of functions into an asymptotic pecking order

A loerarchy?

9.1 A HIERARCHY 427

that includes entries like this:

-x

log

logn 4 logn

+ n’ +

nlogn

4 nn 4

ccn

(Here

and c are arbitrary constants with

< 1 < c.)

All functions listed here, except

go to infinity as n goes to infinity.

Thus when we try to place a new function in this hierarchy, we’re not trying

to determine whether it becomes infinite but rather how fast.

It helps to cultivate an expansive attitude when we’re doing asymptotic

analysis: We should THINK BIG, when imagining a variable that approaches

infinity. For example, the hierarchy says that logn +

n”.ooo’;

this might

seem wrong if we limit our horizons to teeny-tiny numbers like one googol,

n =

10’O”.

For in that case, logn = 100, while

no.ooo’

is only loo.”

1.0233.

But if we go up to a googolplex, n = 1 O”“‘, then logn = 10”’ pales in

comparison with

no.ooo’

10’Oq6.

Even if

is extremely small (smaller than, say,

l/lO'"'oo),

the value

of logn will be much smaller than the value of

n’,

if n is large enough. For

if we set n =

10102k,

where k is so large that

10pk,

we have logn =

10Zk

but n’ 3 1

O'@.

The ratio (logn)/n” therefore approaches zero as n

co.

The hierarchy shown above deals with functions that go to infinity. Often,

however, we’re interested in functions that go to zero, so it’s useful to have

a similar hierarchy for those functions. We get one by taking reciprocals,

because when f(n) and g(n) are never zero we have

1 1

f(n) 4

g(n)

+ f(n)

’

(9.5)

Thus, for example, the following functions (except 1) all go to zero:

-+&+-

nlog

.-&-&+

1 1

-+-

log

log log

Let’s look at a few other functions to see where they fit in. The number

rr(n) of primes less than or equal to n is known to be approximately

n/inn.

Since 1 /ne + 1

/Inn

multiplying by n tells us that

n”’

7r(n)

+ n.

We can in fact generalize (9.4) by noticing, for example, that

na’

(logn)az(loglogn)a3

4 nB’(logn)PZ(loglogn)83

(xl,a2,a3)

(b1,62,p3).

(9.6)

Here ‘(LX’,

012,013)

< (p’,

(32,

(33)’

means lexicographic order (dictionary or-

der); in other words, either a’ < p’, or

0~’

(3’

and

CX~

f12,

a’

(3’

and

0~3

-C

83.

428

ASYMPTOTIC3

How about the functio:n

efi;

where does it live in the hierarchy? We

can answer questions like this by using the rule

which follows in two steps from definition (9.3) by taking logarithms. Conse-

quently

+ f(n) 4 s(n)

==+

eiflnll

e191”ll

And since 1 4 log logn 4

\/logn

logn, we have logn +

+ n6.

When two functions f(n) and g(n) have the same rate of growth, we

write ‘f(n) x g(n)‘. The ofhcial definition is:

f(n)

s(n)

f(n)1

Clg(n)l

and

Is(n)1

Clf(nll,

for some C and for all sufficiently large n.

(9.8)

This holds, for example, if f(n) is constant and g(n) = cos n + arctan

will prove shortly that it

h.olds

whenever f(n) and g(n) are polynomials of

the same degree. There’s

al.so

a stronger relation, defined by the rule

In this case we say that “f(n) is asymptotic to g(n)!’

G. H. Hardy

[148]

introduced an interesting and important concept called

the class of logarithmico-exponential functions, defined recursively as the

smallest family

of functions satisfying the following properties:

The constant function f(n) =

is in

for all real

01.

The identity function f(n) = n is in

If f(n) and g(n) are in

so is f(n)

g(n).

If f(n) is in

so is

efcni.

If f(n) is in

and is “eventually positive,” then lnf(n) is in

A function f(n) is called “eventually positive” if there is an integer

such

that f(n) > 0 whenever n

no.

We can use these rules to show, for example, that f(n) + g(n) is in

whenever f(n) and g(n) are, because f(n) + g(n) = f(n)

(O-g(n)). If f(n)

and g(n) are eventually positive members of

their product f(n) g(n) =

elnf(n)+lnsin) and quotient f(n)/g(n) =

elnf(nlm

lnsini

are in

so are func-

tions like

eilnf(nl,

etc. Hardy proved that every logarithmico-

exponential function is eventually positive, eventually negative, or identically

zero. Therefore the product and quotient of any two C-functions is in

except that we cannot divide by a function that’s identically zero.

9.1 A HIERARCHY 429

wir

durch das

Zeichen 0 (n) eine

GrliSe

ausdrijcken,

deren

Ordnung in

Bezug

auf n

die

Ordnung

von

nicht

iiberschreitet;

sic

wirklich

GIieder

von der

Ordnung n

in sich

enthhlt,

bleibt

bei

dem

bisherigen

SchluDverfahren

dahingestellt.”

Bachmann

[14]

Hardy’s main theorem about logarithmico-exponential functions is that

they form an asymptotic hierarchy: If f(n) and g(n) are any functions in

then either f(n) + g(n), or f(n) + g(n), or f(n) x g(n). In the last case

there is, in fact, a constant a such that

f(n)

as(n).

The proof of Hardy’s theorem is beyond the scope of this book; but it’s nice

to know that the theorem exists, because almost every function we ever need

to deal with is in

In practice, we can generally fit a given function into a

given hierarchy without great difficulty.

9.2

0 NOTATION

A wonderful notational convention for asymptotic analysis was in-

troduced by Paul Bachmann in 1894 and popularized in subsequent years by

Edmund Landau and others. We have seen it in formulas like

lnn+y+O(l/n),

(9.10)

which tells us that the nth harmonic number is equal to the natural logarithm

of n plus Euler’s constant, plus a quantity that is “Big Oh of 1 over n!’ This

last quantity isn’t specified exactly; but whatever it is, the notation claims

that its absolute value is no more than a constant times l/n.

The beauty of O-notation is that it suppresses unimportant detail and

lets us concentrate on salient features: The quantity

O(1

/n) is negligibly

small, if constant multiples of l/n are unimportant.

Furthermore we get to use 0 right in the middle of a formula. If we want

to express

(9.10)

in terms of the notations in Section 9.1, we must transpose

‘Inn +

y’

to the left side and specify a weaker result like

log log n

H,-Inn-y

-X

or a stronger result like

H,-Inn-y

The Big Oh notation allows us to specify an appropriate amount of detail

in place, without transposition.

The idea of imprecisely specified quantities can be made clearer if we

consider some additional examples. We occasionally use the notation

‘

’

stand for something that is either

or -1; we don’t know (or perhaps we

don’t care) which it is, yet we can manipulate it in formulas.

430 ASYMPTOTICS

N. G. de Bruijn begins his book Asymptotic Methods in Analysis by

considering a Big

El1

notation that helps us understand Big Oh. If we write

L(5) for a number whose absolute value is less than 5 (but we don’t say what

the number is), then we ca:n perform certain calculations without knowing

the full truth. For example, we can deduce formulas such as 1 + L(5) = L(6);

L(2) + L(3) = L(5); L(2)L(3) = L(6);

eLc5)

= L(e5); and so on. But we cannot

conclude that L(5)

L(3) = L(2), since the left side might be 4

0. In fact,

the most we can say is L(5)

L(3) = L(8).

Bachmann’s O-notation is similar to L-notation but it’s even less precise:

(01)

stands for a number whose absolute value is at most a constant times

1011.

We don’t say what the number is and we don’t even say what the constant is.

Of course the notion of a

“c:onstant”

is nonsense if there is nothing variable

in the picture, so we use O-notation only in contexts when there’s at least

one quantity (say n) whose value is varying. The formula

f(n) = O(g(n))

for all n

means in this context that there is a constant C such that

(9.11)

If(n)1 6 Clg(n)( for all n;

(9.12)

and when O(g(n)) stands in the middle of a formula it represents a function

f(n) that satisfies (9.12). The values of f(n) are unknown, but we do know

that they aren’t too large. Similarly, de Bruijn’s ‘L(n)’ represents an un-

specified function f(n) whose values satisfy If(n) ( <

In/.

The main difference

between L and 0 is that O-notation involves an unspecified constant C; each

appearance of 0 might involve a different C, but each C is independent of n.

For example, we know that the sum of the first n squares is

$(n+t)(n+l)

in3+tn2+in.

We can write

= O(n3)

because

iin

+ in2 + inI 6 $n13 + SInI + tinI 6 $n31 +

$r31+

iIn

In31

for all integers n. Similarly, we have the more specific formula

On = in3

+O(n2);

we can also be sloppy and throw away information, saying that

III, = O(n’O).

It’s not nonsense,

but it is pointless.

I’ve got a little

list --I’ve got a

little

Iist,

Of annoying terms

and details that

might well be under

ground,

And that never

would be missed

that never would be

missed.

Nothing in the definition of 0 requires us to give a best possible bound.

9.2 0 NOTATION 431

You are the fairest

of your sex,

Let me be your

hero;

I love you as

one over x,

As x approaches

zero.

Positively.

But wait a minute. What if the variable n isn’t an integer? What if we

have a formula like S(x) = $x3 +

ix2

ix,

where x is a real number? Then we

cannot say that S(x) =

0(x3),

because the ratio

S(x)/x3

ix-’

becomes unbounded when x

0. And we cannot say that S(x) = O(x),

because the ratio S(x)/x = $x2 +

becomes unbounded when x

00.

So we apparently can’t use O-notation with S(x).

The answer to this dilemma is that variables used with 0 are generally

subject to side conditions. For example, if we stipulate that

1x1

3 1, or that

x 3

where

is any positive constant, or that x is an integer, then we can

write S(x) =

0(x3).

If we stipulate that

1x1

6 1, or that

1x1

6 c where c is

any positive constant, then we can write S(x) = O(x). The O-notation is

governed by its environment, by constraints on the variables involved.

These constraints are often specified by a limiting relation. For example,

we might say that

f(n)

O(s(n))

n-3

03.

(9.13)

This means that the O-condition is supposed to hold when n is “near”

co;

we don’t care what happens unless n is quite large. Moreover, we don’t

even specify exactly what “near” means; in such cases each appearance of 0

implicitly asserts the existence of two constants C and no, such that

(f(n)1 6 Clg(n)l whenever n >

no.

(9.14)

The values of C and

might be different for each 0, but they do not depend

on n. Similarly, the notation

f(x)

qdxl)

asx+O

means that there exist two constants C and

such that

(f(x)/ 6 Clg(xl(

whenever

1x1

6 E.

(9.15)

The limiting value does not have to be

or 0; we can write

lnz =

z-l+O((z-1)2)

-3 1,

because it can be proved that

Ilnz-z+l/

/z-

112

when

lz-

Our definition of 0 has gradually developed, over a few pages, from some-

thing that seemed pretty obvious to something that seems rather complex; we

now have 0 representing an undefined function and either one or two unspec-

ified constants, depending on the environment. This may seem complicated

enough for any reasonable notation, but it’s still not the whole story! Another

432 ASYMPTOTICS

subtle consideration lurks in the background. Namely, we need to realize that

it’s fine to write

in3 + in2 +

O(n3),

but we should

neueT

write this equality with the sides reversed. Otherwise

we could deduce ridiculous things like n =

from the identities n = 0 (n2)

and

= O(n2). When we work with O-notation and any other formulas

that involve imprecisely specified quantities, we are dealing with one-way

equalities. The right side of an equation does not give more information than

the left side, and it may give less; the right is a “crudification” of the left.

From a strictly formal point of view, the notation 0( g(n)) does not

stand for a single function -f(n), but for the set of all functions f(n) such

that If(n)1 6 Clg(n)l f

or some constant C. An ordinary formula g(n) that

doesn’t involve O-notation stands for the set containing a single function

f(n) = g(n). If S and T are sets of functions of n, the notation S + T stands

for the set of all functions of the form f(n) + g(n), where f(n)

S and

g(n)

T; other notations like S-T, ST, S/T,

es,

In S are defined similarly.

Then an “equation” between such sets of functions is, strictly speaking, a set

inclusion; the ‘=’ sign reall:y means ‘g’. These formal definitions put all of

our 0 manipulations on firm logical ground.

For example, the “equation”

in3 + O(n’) = O(n3)

means that

S2,

where

S-I

is the set of all functions of the form

in3+f1

(n)

such that there exists a constant Cl with

If,(n)/

6 Clln’I, and where

is the set of all functions f.!(n) such that there exists a constant CJ with

Ifz(n)l

6 C21n31.

e can formally prove this “equation” by taking an arbi-

trary element of the left-hand side and showing that it belongs to the right-

hand side: Given in3 + fl In) such that If,(n)1 <

C11n21,

we must prove

that there’s a constant Cl such that l$n3 + fl (n)l 6

C21n31.

The constant

Cl =

+ Cl does the trick,

Isince

In31

for all integers n.

If ‘=’ really means ‘C’, why don’t we use ‘c’ instead of abusing the equals

sign? There are four reasons.

First, tradition. Number theorists started using the equals sign with O-

notation and the practice stuck. It’s sufficiently well established by now that

we cannot hope to get the mathematical community to change.

Second, tradition. Computer people are quite used to seeing equals signs

abused- for years FORTRAN and BASIC programmers have been writing

assignment statements like

“N

= N +

1’.

One more abuse isn’t much.

Third, tradition. We often read ‘=’ as the word ‘is’. For instance we

verbalize the formula

= O(log n) by saying “H sub n is Big Oh of log n!’

“And to auoide the

tediouse repetition

of these woordes:

is equal/e to:

will

sette as

doe often

in woorke use, a

paire of paralleles,

Gemowe

lines of

one lengthe, thus:

bicause

noe

.2.

thynges, can

be moare equal/e.”

-R.

Recorde

12461

“It

obvious that

the sign = is really

the wrong sign

for such relations,

because it suggests

symmetry, and

there is no such

symmetry. . . .

Once this

warning

has been given,

there is, however,

not

much harm

using the sign = ,

and we

shall

main-

tain it,

for no

other

reason than that it

is customary.”

-N. G. de Bruijn

/62]

[Now is a good

time to do

warmup

exercises 3

and

4.)

9.2 0 NOTATION 433

And in English, this ‘is’ is one-way. We say that a bird is an animal, but we

don’t say that an animal is a bird;

“animal” is a crudification of “bird!’

Fourth, for our purposes it’s natural. If we limited our use of O-notation

to situations where it occupies the whole right side of a formula-as in the

harmonic number approximation

= O(log n), or as in the description of

a sorting algorithm’s running time T(n) = O(nlogn) -it wouldn’t matter

whether we used ‘=’ or something else. But when we use O-notation in the

middle of an expression, as we usually do in asymptotic calculations, our

intuition is well satisfied if we think of the equals sign as an equality, and if

we think of something like 0 (1 /n) as a very small quantity.

So we’ll continue to use ‘=I, and we’ll continue to regard O(g(n)) as an

incompletely specified function, knowing that we can always fall back on the

set-theoretic definition if we must.

But we ought to mention one more technicality while we’re picking nits

about definitions: If there are several variables in the environment, O-notation

formally represents sets of functions of two or more variables, not just one.

The domain of each function is every variable that is currently “free” to vary.

This concept can be a bit subtle, because a variable might be defined only

in parts of an expression, when it’s controlled by a

or something similar.

For example, let’s look closely at the equation

(k’ + O(k)) =

in3

+ O(n2), integer n > 0.

(9.16)

k=O

The expression

+ O(k) on the left stands for the set of all two-variable

functions of the form

+ f(k,n) such that there exists a constant C with

]f(k, n)l $ Ck for 0 < k 6 n. The sum of this set of functions, for 0 6 k < n,

is the set of all functions g(n) of the form

$(k’+f(k,n))

~n3+~n2+~n+f(0,n)+f(l,n)+...+f(n,n),

k=O

where f has the stated property. Since we have

(~n2+~n+f(0,n)+f(l,n)+...+f(n,n)l

~n2+/7n2+C.0+C.l

+...+C.n

< n2 + Cl,n2 + n)/2 < (C + l)n2 ,

all such functions g(n) belong to the right-hand side of (9.16); therefore (9.16)

is true.

People sometimes abuse O-notation by assuming that it gives an exact

order of growth; they use it as if it specifies a lower bound as well as an

upper bound. For ex.ample, an algorithm to sort n numbers might be called

434 ASYMPTOTICS

inefficient “because its running time is O(n2)!’ But a running time of 0 (nL)

does not imply that the rrrnning time is not also O(n). There’s another

notation, Big Omega, for lower bounds:

f(n) =

fl(gin))

If(n)1 3 Clg(nil

for some C > 0. (9.17)

We have f(n) =

fI(g(n))

if and only if g(n) = O(f(n)). A sorting algorithm

whose running time is n( n’

)

is inefficient compared with one whose running

time is 0 (n log n) , if n is large enough.

Finally there’s Big Theta, which specifies an exact order of growth:

Since

and

are

umercase

Greek

f(n)

O(g(n))

‘In)

o(g(n))

and

f(n)

n(g(n))

It%ers,

the 0 in

(9.18) O-notation must

be a capital Greek

We have f(n) = @(g(n)) if

<and

only if f(n) x g(n) in the notation we saw

Omicrdn.

After

a//,

Greeks

in-

previously, equation (9.8).

vented asymptotics.

Edmund Landau

[194]

invented a “little oh” notation,

f(n)

o(g(n))

(f(n)1

lElg(n

,)I

for all n 3 no(e) and

for all constants

> 0.

(9.19)

This is essentially the relation f(n

+ g(n) of (9.8). We also have

f(n)

4(n)

f(n)

s(n)

+0(64(n)).

(9.20)

Many authors use

‘0’

in asymptotic formulas, but a more explicit ‘0’ ex-

pression is almost always preferable. For example, the average running time

of a computer method called. “bubblesort” depends on the asymptotic value

of the sum P(n) = ,YF=, k

kl/n’. Elementary asymptotic methods suffice

. .

to prove that P(n)

m?!,

which means that the ratio

P(n)/a

ap-

proaches 1 as n

co. However, the true behavior of P(n) is best understood

by considering the

d@eerence,

P(n)

J7cn/2, not the ratio:

(

P(nllJ7m72

P(n)

0.798

-0.253

10 0.878

-0.484

0.904

-0.538

0.918

-0.561

0.927 -0.575

0.934

-0.585

The numerical evidence in the middle column is not very compelling; it cer-

tainly is far from a dramatic proof that

P(n)/-

approaches 1 rapidly,

9.2 0 NOTATION 435

if at all. But the right-hand column shows that P(n) is very close indeed to

,/%$.

Thus we can characterize the behavior of P(n) much better if we can

derive formulas of the form

P(n)

&72+0(l),

or even sharper estimates like

P(n) =

$ZQ?-

$+0(1/&x)

Stronger methods of asymptotic analysis are needed to prove O-results, but

the additional effort required to learn these stronger methods is amply com-

pensated by the improved understanding that comes with O-bounds.

Moreover, many sorting algorithms have running times of the form

T(n) = Anlgn + Bn + O(logn)

Also ID, the

Dura-

Aame

logarithm.

Notice that

log log log n

is undefined when

n=2.

for some constants A and B. Analyses that stop at T(n)

Anlgn don’t tell

the whole story, and it turns out to be a bad strategy to choose a sorting algo-

rithm based just on its A value. Algorithms with a good ‘A’ often achieve this

at the expense of a bad ‘B’. Since nlgn grows only slightly faster than n, the

algorithm that’s faster asymptotically (the one with a slightly smaller A value)

might be faster only for values of n that never actually arise in practice. Thus,

asymptotic methods that allow us to go past the first term and evaluate B

are necessary if we are to make the right choice of method.

Before we go on to study 0, let’s talk about one more small aspect of

mathematical style. Three different notations for logarithms have been used

in this chapter: lg,

In,

and log. We often use ‘lg’ in connection with computer

methods, because binary logarithms are often relevant in such cases; and

we often use

‘In

in purely mathematical calculations, since the formulas for

natural logarithms are nice and simple. But what about ‘log’? Isn’t this

the “common” base-10 logarithm that students learn in high school-the

“common” logarithm that turns out to be very uncommon in mathematics

and computer science? Yes; and many mathematicians confuse the issue

by using ‘log’ to stand for natural logarithms or binary logarithms. There

is no universal agreement here.

But we can usually breathe a sigh of relief

when a logarithm appears inside O-notation, because 0 ignores multiplicative

constants. There is no difference between O(lgn), O(lnn), and O(logn), as

--+

00;

similarly, there is no difference between 0 (Ig lg n), 0

(In

n), and

O(loglog n). We get to choose whichever we please; and the one with ‘log’

seems friendlier because it is more pronounceable. Therefore we generally

use ‘log’ in all contexts where it improves readability without introducing

ambiguity.

436 ASYMPTOTICS

9.3

0 MANIPULATION

Like any mathematical formalism, the O-notation has rules of ma-

nipulation that free us from the grungy details of its definition. Once we

prove that the rules are correct, using the definition, we can henceforth work

on a higher plane and forget about actually verifying that one set of functions

is contained in another. We don’t even need to calculate the constants C that

The secret of beinn

are implied by each 0, as long as we follow rules that guarantee the existence

a bore is to tell

of such constants.

everything.

Voltaire

For example, we can prove once and for all that

O(n”‘),

when m 6 m’;

O(f(n))

+0(9(n))

O(lf(n)l+

lg(n)l) .

(9.21)

(9.22)

Then we can sayimmediateby that

$n3+in2+in

O(n3)+O(n3)+O(n3)

O(n3), without the laborious calculations in the previous section.

Here are some more rules that follow easily from the definition:

f(n)

O(f(n))

;

c. O(f(n)) = O(f(n)) ,

if c is constant;

O(O(f(n)))

0(+(n))

;

O(f(n))O(g(n))

O(f(n)s(n))

;

O(f(n) s(n)) =

f(n)O(s(n))

(9.23)

(9.24)

(9.25)

(9.26)

(9.27)

Exercise 9 proves (g.22), and the proofs of the others are similar. We can

always replace something of the form on the left by what’s on the right,

regardless of the side conditions on the variable n.

Equations (9.27) and (9.23) allow us to derive the identity O(f(n)2) =

0 (f(n))

This sometimes helps avoid parentheses, since we can write

O(logn)’ instead of

O((logn)2).

Both of these are preferable to

‘O(log2

n)‘, which is ambiguous because some

authors use it to mean ‘O(loglogn)‘.

Can we also write

0 (log n)

’

instead

Iof

O((logn))‘)

(Note: The formula

O(f(n))2

does not

denote the set of

all functions

g(n)’

where g(n) is in

O(f(n)); such

functions g(n)2

cannot be nega-

tive, but the set

O(f(n))’

includes

negative functions.

In genera/, when

S is a set, the no-

tation

stands

for the set of all

No! This is an abuse of notation, since the set of functions l/O(logn) is

products

s’s2

with

neither a subset nor a superset of 0 (1 /log n). We could legitimately substitute

and

in S,

fI(logn)--’ for 0

((logn)-‘),

but this would be awkward. So we’ll restrict our

not for the set of

all

squares

w;th

use of “exponents outside the 0” to constant, positive integer exponents.

S.)

9.3 0 MANIPULATION 437

Power series give us some of the most useful operations of all. If the sum

S(z)

tanz”

n>O

converges absolutely for some complex number

then

S(z)

O(l),

for all

121

/22/.

This is obvious, because

In particular,

S(z)

O(1) as

0, and S(l/n) = O(1) as n

00,

provided

only that

S(z)

converges for at least one

nonzero

value of

We can use this

principle to truncate a power series at any convenient point and estimate the

remainder with 0. For example, not only is

S(z)

= 0( 1

but

S(z)

= a0

+0(z),

S(z)

al2

O(z2)

and so on, because

S(z) =

ukzk

+zm

a,znem

O$k<m

n>m

and the latter sum is 0 (1). Table 438 lists some of the most useful asymp-

totic formulas, half of which are simply based on truncation of power series

according to this rule.

Dirichlet series, which are sums of the form

tka,

ak/k’,

can be truncated

in a similar way: If a Dirichlet series converges absolutely when

can truncate it at any term and get the approximation

ok/k’ + O(m-‘) ,

l<k<m

Remember that

stands for “‘real

part.”

valid for

!.Xz

9%~.

The asymptotic formula for Bernoulli numbers

Table 438 illustrates this principle.

On the other hand, the asymptotic formulas for H,, n!, and

rr(n)

Table 438 are not truncations of convergent series; if we extended them in-

definitely they would diverge for all values of n. This is particularly easy to

see in the case of n(n), since we have already observed in Section 7.3, Ex-

ample 5, that the power series

tk30

k!/

(In

is everywhere divergent. Yet

these truncations of divergent series turn out to be useful approximations.

138

ASYMPTOTICS

Table 438 Asymptotic approximations, valid as n

and

lnn+y+&-A+&

(‘).

(9.28)

?A!-

. (9.29)

B, = 2[n even](-1

)n,/2

’

&(l+2pn+3~n+O(4mn)).

(9.30)

-4

n(n) =

ilntj2

2!n

-+&$+o(&&

(9.31)

(Inni

‘+r+;+~+~+o(r5i.

(9.32)

ln(l+z)

z-f+$-~+0(z5).

(9.33)

= 1

+z+z2+23+t4+0(25).

1-z

(9.34)

(1 +z)a = 1 +cxz+

(;)d+

(;)z3+

(;)24+o(z’l

(9.35)

An asymptotic approximation is said to have absolute error 0( g(n)) if

it has the form f(n)+O(g(n))

w h

ere f(n) doesn’t involve 0. The approxima-

tion has relative error O(g(n)) if it has the form

f(n)(l

+ O(g(n))) where

f(n) doesn’t involve 0. For example, the approximation for

in Table 438

has absolute error

O(n

6);

the approximation for n! has relative error

O(n4).

(The right-hand side of (9.29) doesn’t actually have the required form f(n) x

(1 + O(n “)), but we could rewrite it

dGi

(f)n(l

‘)

(1 + O(nP4))

5 1

840n3

if we wanted to; a similar calculation is the subject of exercise 12.) The

(Relative error

absolute error of this approximation is O(n”

3.5e

~-“).

Absolute error is related

is nice for taking

to the number of correct decimal digits to the right of the decimal point if

reciprocals, because

,,(,

+ 0(c)) =

the 0 term is ignored; relative error corresponds to the number of correct 1

+0(E).)

“significant figures!’

We can use truncation of power series to prove the general laws

ln(l +

O(f(n)))

O(f(n))

if f(n) < 1;

(9.36)

e”‘f’n)l

= 1 +O(f(n)) ,

if f(n) = O(1).

(9.37)

9.3 0 MANIPULATION 439

(Here we assume that n

00;

similar formulas hold for ln( 1 + 0 (f(x)

))

and

e”(f(x)l

as x

0.)

For

example, let

ln(1

+ g(n)) be any function belonging

to the left side of (9.36). Then there are constants C,

no,

and c such that

(g(n)/

CJf(n.)I

< c < 1 , for all n 3 no.

It follows that the infinite sum

ln(1

+ g(n)) =

g(n).

is(n)

+9(n)‘-...)

converges for all n 3

no,

and the parenthesized series is bounded by the

constant 1 +

+c2

+ . . . . This proves

(g.36),

and the proof of

(9.37)

similar. Equations (9.36) and

(g-37)

combine to give the useful formula

+ O(f(n)))“(g(n)) =

O(f(n)g(n))

f~~‘,;l~~

:;tj.

(9.38)

Problem 1: Return to the Wheel of Fortune.

Let’s try our luck now at a few asymptotic problems. In Chapter 3 we

derived equation (3.13) for the number of winning positions in a certain game:

W =

LN/KJ+;K2+$K-3,

K=[mj.

And we promised that an asymptotic version of W would be derived in Chap-

ter 9. Well, here we are in Chapter 9; let’s try to estimate W, as N

03.

The main idea here is to remove the floor brackets, replacing K by

N113

0 (1). Then we can go further and write

K =

N”3(1

+ O(N-“3))

;

this is called “pulling out the large part!’ (We will be using this trick a lot.)

Now we have

N2’3(1

O(N-1’3))2

N2/3(l

+ O(N-‘/3)) =

N2j3

O(N’13)

by (9.38) and

(9.26).

Similarly

LN/KJ

N’P’/3(1

+ O(N-1’3))-1 + O(1)

N2’3(1

+ O(NP”3)) + O(1) =

N2’3

+ O(N”3).

It follows that the number of winning positions is

N2’3

Ol’N”3)

+ ;(N2/3 + O(N”3)) + O(N’j3) +

O(1)

;N2’3

+ O(N”3).

(9.39)

440 ASYMPTOTICS

Notice how the 0 terms absorb one another until only one remains; this is

typical, and it illustrates why O-notation is useful in the middle of a formula.

Problem 2: Perturbation of Stirling’s formula.

Stirling’s approximation for n! is undoubtedly the most famous asymp-

totic formula of all. We will prove it later in this chapter; for now, let’s just

try to get better acquainted with its properties. We can write one version of

the approximation in the form

n! =

J&G

()(

l+~+~+o(n~3)

n-3

00,

(9.40)

for certain constants a and b. Since this holds for all large n, it must also be

asymptotically true when n is replaced by n

(n-l)! =

dm(v)nP1

l+S+

(

O((n-1

lpi))

(9.41)

We know, of course, that (n

l)! = n!/n; hence the right-hand side of this

formula must simplify to the right-hand side of (g.ao), divided by n.

Let us therefore try to simplify (9.41). The first factor becomes tractable

if we pull out the large part:

J271(n-1) =

&(l

-np1)1’2

diik

O(nP3))

Equation (9.35) has been used here.

Similarly we have

n-l

+ O(nP3)

;

-$(l

-n-le2

$+O(np3);

O((n-

l)-")

O(np3(1

-n-1)-3)

= O(nP3),

The only thing in (9.41) that’s slightly tricky to deal with is the factor

l)nm

‘, which equals

nl -1 n-l

(1-n

nn-l

-n

p')n(l

n-l

+ nP2 + O(nP3)) .

9.3 0 MANIPULATION 441

(We are expanding everything out until we get a relative error of O(nP3),

because the relative error of a product is the sum of the relative errors of the

individual factors. All of the O(nP3) terms will coalesce.)

In order to expand (1

nP’)n, we first compute ln(1

nP’ ) and then

form the exponential, enln(‘Pnm’l:

nP’)n = exp(nln(1

-n-l))

exp(n(-nP’

in-’

in3

+ O(nP4)))

= exp(-1

in-’

in2

+ O(nP3))

= exp(-1) .

exp(-in-‘)

exp(-$n2)

. exp(O(nP3))

exp(-1) . (1

in-’

in2

+ O(nP3))

. (1

in2

O(nP4))

. (1 +

O(nP3))

e-l

in-’

$ne2

+ O(nP3)) .

Here we use the notation expz instead of e’, since it allows us to work with

a complicated exponent on the main line of the formula instead of in the

superscript position. We must expand ln(1

-n’)

with absolute error O(ne4)

in order to end with a relative error of O(nP3), because the logarithm is being

multiplied by n.

The right-hand side of (9.41) has now been reduced to

times

n+‘/e”

times a product of several factors:

in-’

AnP2

+ O(nP3))

. (1 +

n-l

-t nP2 + O(nP3))

. (1

in-’

&nP2

O(nP3))

. (1 +

an-’

+ (a + b)nP2 + O(nP3)) .

Multiplying these out and absorbing all asymptotic terms into one O(n-3)

yields

l+an’+(a$-b-&)nP2+O(nP3).

Hmmm; we were hoping to get 1 +

an’

bn2

+ O(nP3), since that’s what

we need to match the right-hand side of (9.40). Has something gone awry?

No, everything is fine; Table 438 tells us that a = A, hence a + b

& = b.

This perturbation argument doesn’t prove the validity of Stirling’s ap-

proximation, but it does prove something: It proves that formula (9.40) can-

not be valid unless a = A. If we had replaced the O(nA3) in (9.40) by

cne3 + O(nP4) and carried out our calculations to a relative error of O(nP4),

we could have deduced that b =

(This is not the easiest way to determine

the values of a and b, but it works.)

442 ASYMPTOTICS

Problem 3: The nth prime number.

Equation (9.31) is an asymptotic formula for n(n), the number of primes

that do not exceed n. If we replace n by p = P,,, the nth prime number, we

have n(p) = n; hence

as n

00. Let us try to “solve” this equation for p; then we will know the

approximate size of the nth prime.

The first step is to simplify the 0 term. If we divide both sides by

p/lnp,

we find that nlnp/p

1; hence

p/lnp

= O(n) and

O(&)

o(i&J

“(&I*

(We have (logp))’ < (logn))’ because p 3 n.)

The second step is to transpose the two sides of (g.42), except for the

0 term. This is legal because of the general rule

+O(f(n))

b, = a,,

+O(f(n))

(9.43)

(Each of these equations follows from the other if we multiply both sides

by -1 and then add a, + b, to both sides.) Hence

n+O(&)

= n(1 +O(l/logn)) ,

lnp

and we have

p =

nlnp(1

+ O(l/logn)) .

(9.44)

This is an “approximate recurrence” for p =

in terms of itself. Our goal

is to change it into an “approximate closed form,” and we can do this by

unfolding the recurrence asymptotically. So let’s try to unfold (9.44).

By taking logarithms of both sides we deduce that

lnp = lnn+lnlnp + O(l/logn) ,

(9.45)

This value can be substituted for lnp in

(g.&,

but we would like to get rid

of all p’s on the right before making the substitution. Somewhere along the

line, that last p must disappear; we can’t get rid of it in the normal way for

recurrences, because (9.44) doesn’t specify initial conditions for small p.

One way to do the job is to start by proving the weaker result p = O(n2).

This follows if we square (9.44) and divide by pn2,

(lnp12

1 +

O(l/logn))

(

9.3 0 MANIPULATION 443

since the right side approaches zero as n

co. OK, we know that p = O(n2);

therefore log p = 0 (log n) and log log p = 0 (log log n). We can now conclude

from (9.45) that

lnp = Inn + O(loglogn)

;

in fact, with this new estimate in hand we can conclude that In In p = In Inn-t

0 (log log n/log n), and (9.45) now yields

lnp = Inn +

lnlnn+

O(loglogn/logn)

And we can plug this into the right-hand side of (g.44), obtaining

p = nlnn+nlnlnn+O(n).

This is the approximate size of the nth prime.

We can refine this estimate by using a better approximation of n(n) in

place of (9.42). The next term of (9.31) tells us that

Get out the scratch

proceeding as before, we obtain the recurrence

paper again, gang.

p = nlnp (1 i- (lnp)

‘)-‘(1

O(l/logn)‘)

(9.46)

which has a relative error of 0( 1 /logn)2 instead of 0( 1

/logn).

Taking loga-

rithms and retaining proper accuracy (but not too much) now yields

lnp =

lnn+lnlnp+0(1/logn)

= Inn

(

lnlnp

Ann

O(l/logn)2)

;

lnlnn

lnlnp =

lnlnn+

Inn

+o(q$y,,

Finally we substitute these results into (9.47) and our answer finds its way

out:

nlnn+nlnlnn-n+n

%+0(C).

b@)

For example, when

‘n

lo6

this estimate comes to 15631363.8 +

O(n/logn);

the millionth prime is actually 15485863. Exercise 21 shows that a still more

accurate approximation to

results if we begin with a still more accurate

approximation to n(n) in place of (9.46).

444 ASYMPTOTICS

Problem 4: A sum from an old final exam.

When Concrete Mathematics was first taught at Stanford University dur-

ing the 1970-1971 term, students were asked for the asymptotic value of the

sum

1 1

+ 1

-+...+-,

+ 2

+ n

with an absolute error of

O(n-‘).

Let’s imagine that we’ve just been given

this problem on a (take-home) final; what is our first instinctive reaction?

No, we don’t panic. Our first reaction is to THINK BIG. If we set n =

lo”‘,

say, and look at the sum, we see that it consists of n terms, each of

which is slightly less than

l/n2;

hence the sum is slightly less than l/n. In

general, we can usually get a decent start on an asymptotic problem by taking

stock of the situation and getting a ballpark estimate of the answer.

Let’s try to improve the rough estimate by pulling out the largest part

of each term. We have

1 1

n2 + k n2(1

+k/n2)

'(1-;+;-$+0(g).

and so it’s natural to try summing all these approximations:

---

n2 + 1 n2 n4

+$-;+o($J

---$+;-;+ogJ

1 1

+ n

;4+$-$+O(-$)

pn;l)

+...

It looks as if we’re getting

n-’

in2

+ O(nP3), based on the sums of

the first two columns; but the calculations are getting hairy.

If we persevere in this approach, we will ultimately reach the goal; but

we won’t bother to sum the other columns, for two reasons: First, the last

column is going to give us terms that are

O(&),

when n/2 6 k 6 n, so we

will have an error of O(nP5); that’s too big, and we will have to include yet

another column in the expansion. Could the exam-giver have been so sadistic?

Do pajamas have

We suspect that there must be a better way. Second, there is indeed a much

buttons?

better way, staring us right in the face.

9.3 0 MANIPULATION 445

Namely, we know a closed form for S,: It’s just H,,z+,,

H,z.

And we

know a good approximation for harmonic numbers, so we just apply it twice:

Hnz+,, = ln(n2 + n)

1 1

2(n2 + n)

12(n2 + n)2

;

(

H,z =

lnn2+y+&

&+O($J.

Now we can pull out large terms and simplify, as we did when looking at

Stirling’s approximation. We have

ln(n2

+n)

= inn’ +ln 1 +

(

lnn’+J--$+&-...;

= ----

+ n

+I-...;

1 1

-1+3-...

(n2

+n)2

So there’s lots of helpful cancellation, and we find

plus terms that are

O(n’).

A bit of arithmetic and we’re home free:

S, = n-1

3-2

inp3 + inp4

&np5 +

An+

o(n-‘).

(9.50)

It would be nice if we could check this answer numerically, as we did

when we derived exact results in earlier chapters. Asymptotic formulas are

harder to verify; an arbitrarily large constant may be hiding in a 0 term,

so any numerical test is inconclusive. But in practice, we have no reason to

believe that an adversary is trying to trap us, so we can assume that the

unknown O-constants are reasonably small. With a pocket calculator we find

that

= & + & + & + & = 0.2170107; and our asymptotic estimate when

n = 4 comes to

$(1+$(-t+

$(-;+f(f

+;(-&

;+))))

= 0.2170125.

If we had made an error of, say, & in the term for ne6, a difference of

would have shown up in the fifth decimal place; so our asymptotic answer is

probably correct.

446 ASYMPTOTICS

Problem 5: An infinite sum.

We turn now to an asymptotic question posed by Solomon Golomb

[122]:

What is the approximate value of

s,,=x

’

k>,

kNn(k)’

’

(9.51)

where N,(k) is the number of digits required to write k in radix n notation?

First let’s try again for a ballpark estimate. The number of digits, N,(k),

is approximately log, k = log k/log n; so the terms of this sum are roughly

(logn)‘/k(log k)‘. Summing on k gives

(logn)’

l/k(log

k)‘, and this

sum converges to a constant value because it can be compared to the integral

O”

dx 1

O”

x(lnx)2

lnx,

=ln2’

Therefore we expect

to be about C(logn)‘, for some constant C.

Hand-wavy analyses like this are useful for orientation, but we need better

estimates to solve the problem. One idea is to express N,,(k) exactly:

N,(k) = Llog,kJ + 1 .

(9.52)

Thus, for example, k has three radix n digits when

6 k < n3, and this

happens precisely when

Llog,

kj = 2. It follows that N,,(k) > log, k, hence

tkal

l/kN,(k)’

+ (logn)’

l/Wgk)‘.

Proceeding as in Problem 1, we can try to write N,(k) = log,, k + 0( 1)

and substitute this into the formula for

S,.

The term represented here by 0 (1)

is always between 0 and 1, and it is about

on the average, so it seems rather

well-behaved. But still, this isn’t a good enough approximation to tell us

about

S,;

it gives us zero significant figures (that is, high relative error) when

k is small, and these are the terms that contribute the most to the sum. We

need a different idea.

The key (as in Problem 4) is to use our manipulative skills to put the

sum into a more tractable form, before we resort to asymptotic estimates. We

can introduce a new variable of summation, m = N,(k):

[n”-’

< k <

n”‘]

km2

k,mZl

9.3 0 MANIPULATION 447

This may look worse than the sum we began with, but it’s actually a step for-

ward, because we have very good approximations for the harmonic numbers.

Still, we hold back and try to simplify some more. No need to rush into

asymptotics. Summation by parts allows us to group the terms for each value

HnmPi

that we need to approximate:

xH,k-,

&).

k21

For example, H,z

is multiplied by 1 /22 and then by -1 /32. (We have used

the fact that H,,o-, =

= 0.)

Now we’re ready to expand the harmonic numbers. Our experience with

estimating (n

1 )! has taught us that it will be easier to estimate

H,,k

than

H,kP1, since the (n”

)‘s

will be messy; therefore we write

HnkP, =

Hnk

= lnnk

+y+

O(h)

klnn+y-&+0(A).

Our sum now reduces to

~(klnn+y-~+o(~))($-~)

kal

Into a Big Oh.

= (1nn)tl

+yE2

t&(n)

O(t3(n2)).

(9.53)

There are four easy pieces left:

El,

X2,

Es(n), and ,Xs(n’).

Let’s do the

,Xx’s

first, since

,X3(n2)

is the 0 term; then we’ll see what

sort of error we’re getting. (There’s no sense carrying out other calculations

with perfect accuracy if they will be absorbed into a 0 anyway.) This sum is

simply a power series,

X3(x)

(j$

&)x-kt

k21

and the series converges when x 3 1 so we can truncate it at any desired point.

If we stop t3(n2) at the term for k = 1, we get I13(n2) = O(nP2); hence (9.53)

has an absolute error of O(ne2). (To decrease this absolute error, we could

use a better approximation to

Hnk;

but O(nP2) is good enough for now.) If

we truncate

,X3(n)

at the term for k = 2, we get

t3(n) =

in-’

+O(nP2);

this is all the accuracy we need.

448 ASYMPTOTICS

We might as well do

now, since it is so easy:

x(&&T)

k>l

This is the telescoping series (1

-;)+(;-$)+($-&)+...

=l.

Finally,

gives us the leading term of S,, the coefficient of Inn in

(9.53):

k($2

&).

k>l

Thisis

(l-i)+(i-$)+(G-&)+...

$+$+$+-

=HE’

=7r2/6.

(If

we hadn’t applied summation by parts earlier, we would have seen directly

that S,

xk3,(lnn)/k2,

because

H,t-,

-H,tmlP1

Inn; so summation by

parts didn’t help us to evaluate the leading term, although it did make some

of our other work easier.)

Now we have evaluated each of the E’s in (g.53), so we can put everything

together and get the answer to

Golomb’s

problem:

S, =

glnn+,-&+0(h),

Notice that this grows more slowly than our original hand-wavy estimate of

C(logn)‘. Sometimes a discrete sum fails to obey a continuous intuition.

Problem 6: Big Phi.

Near the end of Chapter 4, we observed that the number of fractions in

the

Farey

series

3,,

is 1 +

(#J

(n) , where

O(n) =

q(l)

+(p(2)

+...+cP(n);

and we showed in (4.62) that

@(n)

p(k)

ln/k1

+ n/k1 .

k21

(9.55)

Let us now try to estimate

cD(n)

when n is large. (It was sums like this that

led Bachmann to invent O-notation in the first place.)

Thinking BIG tells us that Q(n) will probably be proportional to

n2.

For if the final factor were just

Ln/k]

instead of

+ n/k], we would have

(0(n)( <

xka,

[n/k]’ 6

xk>,(n/k)2 =

$n2,

because the Mobius func-

tion

p(k)

is either -1, 0, or

+l:

The additional ‘1 +

’

in that final factor

adds

xka,

p(k)

Ln/k]

;

but this is zero for k > n, so it cannot be more than

nH,

= O(nlog n) in absolute value.

9.3 0 MANIPULATION 449

This preliminary analysis indicates that we’ll find it advantageous to

write

‘(n)

;fP(k((;)

+0(1))2

;fp(k)((;)2+o(;))

k=l

;&‘i(;)l+fo(;)

k=l k=l

ifIdk)(E)l

+ O(nlogn)

k=l

This removes the floors; the remaining problem is to evaluate the unfloored

sum

x.L,

p(k)n2,/k2 with an accuracy of O(nlogn); in other words, we

want to evaluate ,Fi’=,

p(k)l/k’

with an accuracy of

O(n-’

logn). But that’s

easy; we can simply run the sum all the way up to k = cq because the newly

added terms are

k>n

O(g2)

O(&x.&)

k>n

(

kJA

-t,)

o(A).

We proved in (7.88) that

tk>,

F(k)/k’

l/<(z).

Hence

tk>,

k(k)/k’

‘/(tk>l

1 /k2) = 6/7r2, and we have our answer:

CD(n)

$n2

+ O(nlogn).

(9.56)

9.4

TWO ASYMPTOTIC TRICKS

Now that we have some facility with 0 manipulations, let’s look at

what we’ve done from a slightly higher perspective. Then we’ll have some

important weapons in our asymptotic arsenal, when we need to do battle

with tougher problems.

nick 1: Boots trapping.

When we estimated the nth prime

in Problem 3 of Section 9.3, we

solved an asymptotic recurrence of the form

nlnP,(l

+ O(l/logn)) .

We proved that

= nln n + O(n) by first using the recurrence to show

the weaker result O(n2). This is a special case of a general method called

bootstrapping, in which we solve a recurrence asymptotically by starting with

450

ASYMPTOTIC3

a rough estimate and plugging it into the recurrence; in this way we can often

derive better and better estimates, “pulling ourselves up by our bootstraps.”

Here’s another problem that illustrates bootstrapping nicely: What is the

asymptotic value of the coefficient

g,,

[zn]

G(z) in the generating function

G(z)

exp(t

k>l

as n

oo?

If we differentiate this equation with respect to z, we find

G’(z) =

ngnznpl =

G(z)

;

n=O

k>l

equating coefficients of zn-’ on both sides gives the recurrence

O<k<n

b@)

Our problem is equivalent to finding an asymptotic formula for the solution

to (g.58), with the initial condition

= 1. The first few values

n 01234

, ,

107

641 51103

4 36 288

2400 259200

don’t reveal much of a pattern, and the integer sequence (n!2g,) doesn’t

appear in Sloane’s Handbook

[270];

therefore a closed form for

seems out

of the question, and asymptotic information is probably the best we can hope

to derive.

Our first handle on this problem is the observation that 0 <

6 1 for

all n 3 0; this is easy to prove by induction. So we have a start:

O(1)

’

This equation can, in fact, be used to “prime the pump” for a bootstrapping

operation: Plugging it in on the right of (9.58) yields

ng,

O(1)

H,O(l)

= O(logn);

O<k<nn-k

hence we have

log n

on’

(

for n > 1.

9.4 TWO ASYMPTOTIC TRICKS 451

And we can bootstrap yet again:

O(U

logk)/k)

n=-

n-k

O<k<n

O(logn)

t+o<&Cnk(n-k)

;

o<&<n(;

--&)

O(longn)

= k +

~H,~,O(logn)

kO(logn)‘,

obtaining

logn

ok>

(9.59)

Will this go on forever? Perhaps we’ll have

g,,

O(n’

logn)m for all m.

Actually no; we have just reached a point of diminishing returns. The

next attempt at bootstrapping involves the sum

O<k<n

k2(n

’

(&‘&+nz(,‘-k))

O<k<n

H(2)

n-,

$%I

which is

n(n-‘);

so we cannot get an estimate for

g,,

that falls below

n(n2).

In fact, we now know enough about

g,,

to apply our old trick of pulling

out the largest part:

Obk<n

gk-;tgk+;

k20

k3n

O<k<n

n-k

(9.60)

The first sum here is G(1) = exp(f +

...)

en2/6,

because G(z)

converges for all

Iz/

6 1. The second sum is the tail of the first; we can get an

upper bound by using (9.59):

tgk

o(+$)

o(““g,“‘2).

k2n

k>n

452 ASYMPTOTICS

This last estimate follows because, for example,

k>n

(Exercise 54 discusses a more general way to estimate such tails.)

The third sum in (9.60) is

by an argument that’s already familiar. So (9.60) proves that

(log

n/n)3

Finally, we can feed this formula back into the recurrence, bootstrapping once

more; the result is

en2/b

+ O(logn/n3)

(Exercise 23 peeks inside the remaining 0 term.)

Trick

2: Trading

tails.

We derived (9.62) in somewhat the same way we derived the asymptotic

value (9.56) of O(n): In both cases we started with a finite sum but got an

asymptotic value by considering an infinite sum. We couldn’t simply get the

infinite sum by introducing 0 into the summand; we had to be careful to use

one approach when k was small and another when k was large.

Those derivations were special cases of an important three-step asymp-

(This impor-

totic summation method we will now discuss in greater generality. Whenever

tant

method

waS

we want to estimate the value of

(n), we can try the following approach:

pioneered by

Lap/ace [195

‘1.)

First break the sum into two disjoint ranges,

and

T,,.

The summation

over

should be the “dominant” part, in the sense that it includes

enough terms to determine the significant digits of the sum, when n is

large. The summation over the other range

T,,

should be just the “tail”

end, which contributes little to the overall total.

2 Find an asymptotic estimate

ak(n)

bk(n)

O(ck(n))

that is valid when k

D,. The 0 bound need not hold when k

T,.

9.4 TWO ASYMPTOTIC TRICKS 153

at each of the following three sums is small:3 Now prove th

L(n)

ok(n); tb(n)

‘Jk(n)

;

MT,

kET,

(n)

(ck(n)l.

(9W

If all three steps can be completed successfully, we have a good estimate:

ak(n)

bk(n)

o(L(n))

O(xb(n))

o(L(n))

kED,uT, kED,uT,

Here’s why. We can “chop

off”

the tail of the given sum, getting a good

estimate in the range

where a good estimate is necessary:

ak(n)

@k(n)

O(ck(n)))

bk(n)

o&(n)).

G-D, kCD,

ND,

And we can replace the tail with another one, even though the new tail might

be a terrible approximation to the old, because the tails don’t really matter:

Asymptotics is

the art of knowing

where to be sloppy

and where to be

precise.

ak(n)

@k(n)

bk(n)

ak(n))

&T, MT,

h(n)

O(xb(n))

o&,(n)).

MT,

When we evaluated the sum in (g.6o), for example, we had

ak(n)

[06k<nlgk/(n-kk),

h(n)

Sk/n,

ck(n)

kgk/n(n-k);

the ranges of summation were

{O,l,...,

n-l},

T,,

{n,n+l,...};

and we found that

x,(n) =

Lb(n)

= o((logn)2/n2), xc(n) = o((logn)3/n2).

This led to (9.61).

Similarly, when we estimated

0(n)

in (9.55) we had

ak(n)

v(k)

[n/k]

Ll+n/k]

bk(n)

dk)n2/k2

ck(n)

n/k;

= {1,2

,...,

n},

T,,

{n+l,n+2,...}.

We derived (9.56) by observing that

E,(n)

= 0, xb(n) = O(n), and L,(n) =

O(nlogn).

454 ASYMPTOTICS

Here’s another example where tail switching is effective. (Unlike our

Also, horses switch

previous examples, this one illustrates the trick in its full generality, with

their

when

,X,(n) # 0.) We seek the asymptotic value of

feeding time ap-

proaches.

The big contributions to this sum occur when k is small, because of the k! in

the denominator. In this range we have

ln(n+2k)

lnn+c-2+0(s)

b64

We can prove that this estimate holds for 0 6 k <

Llg

n] , since the original

terms that have been truncated with 0 are bounded by the convergent series

2km

23k

mnm

-...

m33

(In this range, 2”/n 6 2L1snlP1/n 6

i.)

Therefore we can apply the three-step method just described, with

ok(n) = ln(n + 2k)/k! ,

bk(n) = (lnn + 2”/n

4k/2n2)/k!,

ck(n) = gk/n3k!;

{O,l,...,

[lgn]

-l},

T,,

= {LlgnJ,[lgnj

+l,...}.

All we have to do is find good bounds on the three

t’s

in (g.63), and we’ll

know

that

tk>(,

ak(n)

tk>‘,

bk(n).

The error we have committed in the dominant part of the sum, L,(n) =

keD, gk/n3k!, is obviously bounded by tk>O gk/n3k! = e8/n3, so it can be

replaced by O(nP3). The new tail error is

lnn+2k+4k

Llg

lnn+2lknJ

+4llsnl

9.4 TWO ASYMPTOTIC TRICKS 455

“We may not be big,

Since Llgnj ! grows faster than any power of n, this minuscule error is

over-

but we’re small.”

whelmed by X,(n)

O(nP3). The error that comes from the original tail,

is smaller yet.

Finally, it’s easy to sum

k20

bk(n) in closed form, and we have obtained

the desired asymptotic formula:

ln(n + 2k)

k’

k20

’

elnnt---+0

The method we’ve used makes it clear that, in fact,

k20

ln(n

2k)

m-1

elnn+

~(-l)k+l&+O(-&)j

k=l

(w%)

(9.66)

for any fixed m > 0. (This is a truncation of a series that diverges for all

fixed n if we let m -+ co.)

There’s only one flaw in our solution: We were too cautious. We de-

rived (9.64) on the

iassumption

that k < [lgn], but exercise 53 proves that

the stated estimate is actually valid for all values of k. If we had known

the stronger general result, we wouldn’t have had to use the two-tail trick;

we could have gone directly to the final formula! But later we’ll encounter

problems where exchange of tails is the only decent approach available.

9.5 EULER’S SUMMATION FORMULA

And now for our next trick-which is, in fact, the last important

technique that will be discussed in this book-we turn to a general method of

approximating sums that was first published by Leonhard Euler

[82]

in 1732.

(The idea is sometimes also associated with the name of Colin Maclaurin,

a professor of mathematics at Edinburgh who discovered it independently a

short time later

[211,

page

3051.)

Here’s the formula:

f(k)

1”

f(x)dx

L$f+‘)(x)ib

+ R,,

b67)

a<k<b

where

(--l)m+’

&(1x))

fcmi(x) dx

integers a < b; (g 68)

integer m 3 1.

’

456 ASYMPTOTICS

On the left is a typical sum that we might want to evaluate. On the right is

another expression for that sum, involving integrals and derivatives. If f(x) is

a sufficiently “smooth” function, it will have m derivatives f’(x), . . . ,

f(“)

(x),

and this formula turns out to be an identity. The right-hand side is often an

excellent approximation to the sum on the left, in the sense that the remain-

der

is often small. For example, we’ll see that Stirling’s approximation

for n! is a consequence of Euler’s summation formula; so is our asymptotic

approximation for the harmonic number H,.

The numbers

in (9.67) are the Bernoulli numbers that we met in

Chapter 6; the function B,({x}) in (9.68) is the Bernoulli polynomial that we

met in Chapter 7. The notation {x} stands for the fractional part x

Lx],

in Chapter 3. Euler’s summation formula sort of brings everything together.

Let’s recall the values of small Bernoulli numbers, since it’s always handy

to have them listed near Euler’s general formula:

B, =

-5,

-&-,

= &,

-$,;

= Bs =

= B,, = . . . =

Jakob Bernoulli discovered these numbers when studying the sums of powers

of integers, and Euler’s formula explains why: If we set f(x) = x”-‘ , we have

f’“‘(x)

= 0; hence

,,,

= 0, and (9.67) reduces to

aSk<b

Bk. (bm-k

ampk)

For example, when m = 3 we have our favorite example of summation:

i((i)Bon3+(f)Bln’+(z)B2n)

T--T+:

OSk<n

(This is the last time we shall derive this famous formula in this book.) All

good things

Before we prove Euler’s formula, let’s look at a high-level reason (due

to Lagrange

[192])

why such a formula ought to exist. Chapter 2 defines the

~nu~~c’me

’

difference operator A and explains that

is the inverse of A, just as J is the

inverse of the derivative operator D. We can express A in terms of D using

Taylor’s formula as follows:

f(X

+ E)

f(x) +

. . .

9.5 EULER’S SUMMATION FORMULA 457

Setting

= 1 tells us that

Af(x) = f(x+ 1) -f(x)

f/(x)/l!

+ f”(X)/2! + f”‘(X)/3! +

“.

(D/l!+D2/2!+D3/3!+...)f(x)

(eD-l)f(x).

b69)

Here

stands for the differential operation 1 + D/l ! + D2/2! + D3/3! + . . . .

Since A =

the inverse operator

= l/A should be l/(eD

1); and

we know from Table 337 that z/(e’

1) =

Bk.zk/k!

is a power series

involving Bernoulli numbers. Thus

~+!?+$D+$D~+...

J+&$Dkpl.

(9.70)

Applying this operator equation to f(x) and attaching limits yields

~;f(x)hx

f(x) dx +

sf’kpl’(x)

k>l

(9.71)

which is exactly Euler’s summation formula (9.67) without the remainder

term. (Euler did not, in fact, consider the remainder, nor did anybody else

until S. D. Poisson

[:236]

published an important memoir about approximate

summation in 1823. The remainder term is important, because the infinite

sum

xk>,

(Bk/k!)fCk--‘)(x)li often diverges. Our derivation of (9.71) has been

purely formal, without regard to convergence.)

Now let’s prove (g.67),

with the remainder included. It suffices to prove

the case a = 0 and b =

namely

f(0) = f(x) tix +

wyx)~’

(-l)m

’

Bm(x)

(ml

k=,

o 0

m! f (x)

’

because we can then replace f(x) by f (x +

for any integer 1, getting

lfl

f(l) =

f(x)dx+f

~f(kpl)(x)lL+‘-

(-l)m

k=,

I+’

Bm()’

f’“’

(x)

The general formula (9.67) is just the sum of this identity over the range

a 6

< b, because intermediate terms telescope nicely.

The proof when a = 0 and b = 1 is by induction on m, starting with

m= 1:

f(0) =

J’f(x)dx-f(f(l)-f(o))+J’(x-;)f’(x)dx.

458 ASYMPTOTICS

(The Bernoulli polynomial E&(x) is defined by the equation

B,(x)

(y)Boxm+

(~)B,x~-’

+...+

(~)B,x~

(9.72)

in general, hence

(x) = x

in particular.) In other words, we want to

prove that

f(O) +

f(l)

/;f(x)dx+l:jx-;)f’lx)dx.

But this is just a special case of the formula

u(xMx)

11,

u(x) dv(x) +

4x1

du(x)

(9.73)

0 0

for integration by parts, with u(x) = f(x) and v(x) = x

Hence the case

= 1 is easy.

To pass from m

1 to m and complete the induction when m >

need to show that

R,-l

= (B,/m!)f(mP1’(~)l~ +

R,,

namely that

This reduces to the equation

(-l)mBmf(mpli (x)1’

= m

J’B,-

(x)Grnp’)(x)

JIB,,,(xlGml(x)

dx.

0 0

Once again (9.73) applies to these two integrals, with u(x) = f(“-

‘l(x)

and

Will the authors

v(x) = B,(x), because the

d.erivative

of the Bernoulli polynomial (9.72) is

never get serious?

mB,-l(x).

(9.74)

(The absorption identity (5.7) was useful here.) Therefore the required for-

mula will hold if and only if

(-l)“‘B,,,f(“~‘)

(x)1;

B,(x)f’mpl)(x)l;.

9.5 EULER’S SUMMATION FORMULA 459

In other words, we need to have

(-l)mBm

B,,(l)

B,(O),

for

(9.75)

This is a bit embarrassing, because B,(O) is obviously equal to

B,,

not

(-l)mB,.

But there’s no problem really, because m > 1; we know that

is zero when m is odd. (Still, that was a close call.)

To complete the proof of Euler’s summation formula we need to show

that B,,,(l) = B,(O), which is the same as saying that

for m

But this is just the definition of Bernoulli numbers,

(6.7g),

so we’re done.

The identity

B&(x)

= mBm-l (x) implies that

Bm(x) dx =

,+1(l)

Bm+l(O)

m+l

and we know now that this integral is zero when m 3

Hence the remainder

term in Euler’s formula,

(-‘);+’

i”

Bm((x))f(“‘)(x)

dx,

multiplies f’“)(x) by a function

({x}) whose average value is zero. This

means that

has a reasonable chance of being small.

Let’s look more closely at B,(x) for 0 6 x 6 1, since B,(x) governs the

behavior of R,.

Here are the graphs for

B,(x)

for the first twelve values of m:

m=2 m=3

Bm(x)

W-

4+m(X)

BS+m(X)

m=4

Although BJ (x) through

Bg(x)

are quite small, the Bernoulli polynomials

and numbers ultimately get quite large. Fortunately

has a compensating

factor 1

/m!,

which helps to calm things down.

460 ASYMPTOTICS

The graph of B,(x) begins to look very much like a sine wave when

m > 3; exercise 58 proves that B,(x) can in fact be well approximated by a

negative multiple of

cos(27rx

inm),

with relative error

l/2”.

In general,

Bdk+l

(x) is negative for 0 < x <

and positive for

< x < 1.

Therefore its integral,

Bdk+~

(x)/(4k+2),

decreases for 0 < x <

and increases

for

< x < 1. Moreover, we have

bk+l(l

-&+I

(X)

for 0 < x <

and it follows that

bk+2(1

-X)

bk+2(x),

for 0 < x <

The constant term

Bdk+2

causes the integral

l&k+l(x) dx to be zero; hence

B4k+2

> 0. The integral of

Bak+Z(X)

is Bdk+A(x)/(4k+3), which must therefore

be positive when 0 < x <

and negative when i < x < 1; furthermore

B4k+3

( 1

x) = -B4k+3 (x) , so B4k+j (x) has the properties stated

for

i&k+1 (x),

but negated. Therefore

B4k

+4(x) has the properties stated for BJ~+z(x), but

negated. Therefore B4k+s(x) has the properties stated for

B4k+l

(x); we have

completed a cycle that establishes the stated properties inductively for all k.

According to this analysis, the maximum value of Blm(x) must occur

either at x = 0 or at x =

Exercise 17 proves that

BZm(;) =

(21mm2”’

1)‘B2,,,;

(9.76)

hence we have

(bn(b4I

IBzml.

(9.77)

This can be used to establish. a useful upper bound on the remainder in Euler’s

summation formula, because we know from (6.89) that

IBZllJ

(2m)!

when m > 0.

Therefore we can rewrite Euler’s formula (9.67) as follows:

f(k)

J”

a<k<b

o((2n)~~~)

Jbpyxq

dx.

(9.78)

For example, if f(x) = ex, all derivatives are the same and this formula tells

us that

taSkCb

(eb

ea)(l

+ B2/2!

B4/4! +

...

B&(2m)!)

9.5 EULER’S SUMMATION FORMULA 461

0((2n)-2”). Of

course, we know that this sum is actually a geometric series,

equal to

(eb

e”)/(e

1) =

(eb

ea)

xkSO

Bk/k!.

f(2m)(x)

for a < x < b, the integral

lf(2")(x)l

dx is just

f(2m-1)(x)li,

so we have

‘R2m’

(237q!

2!.?+yX)~~

;

in other words, the remainder is bounded by the magnitude of the final term

(the term just before the remainder), in this case. We can give an even better

estimate if we know that

f(2m+2)(x)

> 0 and

f(2m+41(x)

for a 6 x 6 b.

(g-79)

For it turns out that this implies the relation

B2m+2

R2,,,

8,---

(2m +

2)!

f(2m+')(x)l;,

forsomeO<0,<1;

b8o)

in other words, the remainder will then lie between 0 and the first discarded

term in

(9.78)

-the term that would follow the final term if we increased m.

Here’s the proof: Euler’s summation formula is valid for all m, and

Bz,,,+I

= 0 when m > 0; hence

Rz,,,

Rz,+,,

and the first discarded term

must be

R2,+2.

We therefore want to show that

Rzm

lies between 0 and

R2m

Rzm+2;

and

this is true if and only if

Rz,,

and

R2,,,+2

have opposite signs. We claim that

f(Zm+2)(x)

0 for a < x 6 b implies (-l)“‘Rz,,, 3 0.

(9.81)

This, together with

(g.Tg),

will prove that R

z,,,

and

R~,,,+z

have opposite signs,

so the proof of (9.80) will be complete.

It’s not difficult to prove

(9.81)

if we recall the definition of

Rzm+l

and

the facts we proved about the graph of

Bzm+l

(x). Namely, we have

R2m = R2,.+1 =

B2m+1

6”)

f(h+l)

(x)

(2m+ l)!

and

f(2m+')(x)

is increasing because its derivative

f12m+2)(~)

is positive. (More

precisely,

(2m+'

(x) is nondecreasing because its derivative is nonnegative.)

The graph of

Bz,,,+r

({x}) looks like (-1)

m+’

times a sine wave, so it is geo-

metrically obvious that the second half of each sine wave is more influential

than the first half when it is multiplied by an increasing function. This makes

(-l)mR2,+~

0, as desired. Exercise 16 proves the result formally.

462 ASYMPTOTICS

9.6 FINAL SUM:MATIONS

Now comes the summing up, as we prepare to conclude this book.

We will apply Euler’s summation formula to some interesting and important

examples.

Summation

This one jis too easy.

But first we will consider an interesting unimportant example, namely

a sum that we already know how to do. Let’s see what Euler’s summation

formula tells us if we apply

i.t

to the telescoping sum

sn=

EL=:

l<k<nkck+')

---

I<k<n

k+l

l-1

n’

\ \

It can’t hurt to embark on our first serious application of Euler’s formula with

the asymptotic equivalent of training wheels.

We might as well start by writing the function f(x) = 1 /x(x+ 1) in partial

fraction form,

f(x) =

;

--$

since this makes it easier to integrate and differentiate. Indeed, we have

f'(x) =

-l/x2

+ l/(x +

1)2

and f"(x) =

2/x3

-2/(x +

l)3;

in general

f(k)(,)

(-l)“k’

-!w

’

Jxk+I

(x+l)kf'

for k > 0.

Furthermore

f(x)dx

lnx-ln(x+l))y

ln--$.

Plugging this into the summation formula (9.67) gives

= In

&,)k!f

v$-

k=l

(

’

(n+l)k

-1

+R,(n),

wheel

-/~B,,({xl)(&-

(x+~)m+,)dx.

For example, the right-hand side when m = 4 is

2n 1 1 1 1

(

---.--

1 1 1

~--

--~--

n+l

n-t1

(

(n+1)2

120

(

;;7

R4(n)

9.6 FINAL SUMMATIONS 463

This is kind of a mess; it certainly doesn’t look like the real answer 1

n-l.

But let’s keep going anyway, to see what we’ve got. We know how to expand

the right-hand terms in negative powers of n up to, say,

O(n5):

In n

-n-l

Lgp2

in-3

$np4 + o(ne5)

;

n+l

-1

,-2

np3

+ O(np5);

___

(n-t

1)2

nm2

+ 3n

+ O(np5)

;

np4

O(n5)

Therefore the terms on the right of our approximation add up to

ln2 + t + &

+ (-1

t)n-’

& + &)np2

+(-+-&)n-3

(1;

+ & +

&)nm4

+ R4(n)

ln2+$$-n-‘+R4(n)+O(n~5).

The coefficients of ne2, np3, and

nm4

cancel nicely, as they should.

If all were well with the world, we would be able to show that

R,+(n)

asymptotically small, maybe

O(n5),

and we would have an approximation

to the sum. But we can’t possibly show this, because we happen to know that

the correct constant term is

not ln2 +

(which is approximately 0.9978).

So R4(n) is actually equal to

ln2 +

O(n4),

but Euler’s summation

formula doesn’t tell us this.

In other words, we lose.

One way to try fixing things is to notice that the constant terms in the

approximation form a. pattern, if we let m get larger and larger:

ln2-tB,+t.~B2-~.~B3+~.~B4--.~B5+...

Perhaps we can show that this series approaches 1 as the number of terms

becomes infinite? But no; the Bernoulli numbers get very large. For example,

f322

@$$

> 6192; therefore

IR22(n)l

will be much larger than 1

(n)

We lose totally.

There is a way out, however, and this escape route will turn out to be

important in other applications of Euler’s formula. The key is to notice that

R4(n) approaches a definite limit as n

00:

lim R4(n) =

-jyB4({x))(&&)

R4(m)

n-02

464 ASYMPTOTICS

The integral

B4({x})f’“:(x) dx will exist whenever

f’“‘(x)

0(x

‘)

00, and in this case

f14)

(x) surely qualifies. Moreover, we have

R4(n) =

Rl(m!+~~Bl({x:)(~-~)dn

R4(00)

O(/“xp6

)dx =

R~(cw)

+ O(nP5).

Thus we have used Euler’s summation formula to prove that

= ln2 -t

-n-l

+ R4(00) +

O(n5)

= C

I-I-’

+ O(ne5)

for some constant C. We do not know what the constant is-some other

method must be used to establish it -but Euler’s summation formula is able

to let us deduce that the

co,nstant

exists.

Suppose we had chosen a much larger value of m. Then the same rea-

soning would tell us that

R,(n) = R,(m) + O(nPmP’),

and we would have the formula

<k<n

k(k+l

)

T-C’

+ c2nP2 + cjnP3 + . . . +

c,n~~

+ O(nPmP’ )

for certain constants

~2,

~3,

. . . . We know that the c’s happen to be zero

in this case; but let’s prove it, just to restore some of our confidence (in

Euler’s formula if not in ourselves). The term In

contributes (-1

)“/m

to cm; the term

(-l)m+’

(Bm/m)nPm contributes

(-l)“+‘B,/m;

and the

term

(-l)k(Bk/k)(n+

l)pk contributes

(-l)m(F::)BJk.

Therefore

(-‘)ym

A-%

(-1);

k=l

k-1

-1

m m

;(I-B,+B,(l)-1).

Sure enough, it’s zero, when m > 1. We have proved that

= C

-n-l

+ O(nPmP’),

for all m > 1 .

(9.82)

This is not enough to prove that the sum is exactly equal to C

’

;

the

actual value may be C

n’

2-”

or something. But Euler’s summation

9.6 FINAL SUMMATIONS 465

formula does give us

O(n

mP1

) for arbitrarily large m, even though we haven’t

evaluated any remainders explicitly.

Summation 1, again: Recapitulation and generalization.

Before we leave our training wheels, let’s review what we just did from

a somewhat higher perspective. We began with a sum

S, =

f(k)

l<k<n

and we used Euler’s

isummation

formula to write

S, =

F(n)

F(‘I

1-t

c(Tkin)

Tk(l

R,(n),

k=l

(9.83)

where F(x) was

f(x) dx and where

(x)

was a certain term involving Bk and

f(kmm’)(~).

We also noticed that there was a constant c such that

fcm)(x)

0(x’-“)

as x

00,

for all large m.

(Namely, f(k) was

l/k(k+

1); F(x) was ln(x/(x+ 1));

Tk(x)

was (-l)k+’ x

(Bk/k)(x-k

+ 1)-k);

and c was -2.) For all large enough values of m,

this implied that the remainders had a small tail,

R,!,,(n) = R,(Lx)

R,(n)

(-y+’

O”

Bm(bl)

,f’““(x)

dx =

O(nc+‘Pm).

(9.84)

Therefore we were able to conclude that there exists a constant C such that

S, = F(n) + C

-t

Tk(n)

R,/,,(n).

k=l

(9.85)

(Notice that C nicely absorbed the

Tk(

1) terms, which were a nuisance.)

We can save ourselves unnecessary work in future problems by simply

asserting the

existenc.e

of C whenever R,,(m) exists.

Now let’s suppose that

f(2m+21(x)

3 0 and

f(2m+4)(~)

3 0 for 1 6 x 6 n.

We have proved that this implies a simple bound (9.80) on the remainder,

b,(n)

%,,,(Tzm+2(n)

Tzm+2(l

where

8,,,

lies somewhere between 0 and

But we don’t really want bounds

that involve Rz,(n) and

T2,,,+2(

1); after all, we got rid of

Tk(

1) when we

introduced the constant C. What we really want is a bound like

-%,,(n)

hnTzm+2(n),

466

ASYMPTOTIC3

where 0 <

a,,,,

< 1; this will allow us to conclude from (9.85) that

S, = F(n) +

(n)

Tzk(n)

Gm,nT2m+2(n)

k=l

(9.86)

hence the remainder will truly be between zero and the first discarded term.

A slight modification of our previous argument will patch things up per-

fectly. Let us assume that

f(2m+2'(x)

0 and

fc2m+4)(x)

0, as x

0~).

(9.87)

The right-hand side of (9.85) is just like the negative of the right-hand side of

Euler’s summation formula (9.67) with a = n and b = 00, as far as remainder

terms are concerned, and successive remainders are generated by induction

on m. Therefore our previous argument can be applied.

Summation 2: Harmonic numbers harmonized.

Now that we’ve learned so much from a trivial (but safe) example, we can

readily do a nontrivial one. Let us use Euler’s summation formula to derive

the approximation for

that we have been claiming for some time.

In this case, f(x) = l/x. We already know about the integral and deriva-

tives of f, because of Summation 1; also

f(ml(x)

O(xpmp')

as x

00.

Therefore we can immediately plug into formula (9.85):

l<k<n

Inn + C +

Bin-’

2kn2k

R&(n),

k=l

for some constant C. The

;sum

on the left is Hn-lr not

H,;

but it’s more

convenient to work with H,-~l and to add 1

later, than to mess around with

(n +

1)'s

on the right-hand side. The

Bin-l

will then become (B, + 1

)n-'

1/(2n). Let us call the constant y instead of C, since Euler’s constant y is,

in fact, defined to be

lim,,,,

(H,

Inn).

The remainder term can be estimated nicely by the theory we developed

a minute ago, because

f(2")(x)

= (2m)!/x2”‘+’ 3 0 for all x > 0. Therefore

(9.86) tells us that

(9.88)

where

0,,,

is some fraction between 0 and

This is the general formula

whose first few terms are listed in Table 438. For example, when m = 2 we get

lnn+y+K-m

+ 120n4

02 n

-252n6*

(9.89)

9.6 FINAL SUMMATIONS 467

This equation, incidentally, gives us a good approximation to y even when

n = 2:

y =

Hz-ln2--i

+&-&,+e

0.577165...+~,

where

is between zero and

If we take n =

lo4

and m = 250, we get

the value of y correct to 1271 decimal places, beginning thus

[171]:

y =

0.57721566490153286060651209008240243...

(9.90)

But Euler’s constant appears also in other formulas that allow it to be eval-

uated even more efficiently

[282].

Summation 3: Stirling’s approximation.

If f(x) = In x, we have f'(x) = 1

/x,

so we can evaluate the sum of

logarithms using almost the same calculations as we did when summing re-

ciprocals. Euler’s summation formula yields

$k<n

Ink =

nl.nn-n+o-F

B2k

B2m+2

k=,

2k(2k-l)nzk-’

’

(Pm’n

(2m+2)(2m+l)n2m+l

where u is a certain constant, “Stirling’s constant,” and 0 <

(P~,~

(In this

case

f(2")(x)

is negative, not positive; but we can still say that the remainder

is governed by the first discarded term, because we could have started with

f(x) =

-In

x instead of f(x) = lnx.) Adding Inn to both sides gives

(P2,n

Inn! =

nlnn--n+F+o+&--

___

360n3

’

1260n5

(9.91)

Heisenberg may

have been here.

when m = 2. And we can get the approximation in Table 438 by taking ‘exp’

of both sides. (The value of

turns out to be

fi,

but we aren’t quite ready

to derive that formula. In fact, Stirling didn’t discover the closed form for

until several years after de Moivre

[64]

had proved that the constant exists.)

If m is fixed and n

00, the general formula gives a better and better

approximation to Inn! in the sense of absolute error, hence it gives a better

and better approximation to n! in the sense of relative error. But if n is fixed

and m increases, the error bound

IB2,+21/(2m

+ 2)(2m + 1

)n2”‘+’

decreases

to a certain point

anld

then begins to increase. Therefore the approximation

reaches a point beyond which a sort of uncertainty principle limits the amount

by which n! can be approximated.

468 ASYMPTOTICS

In Chapter 5, equation

(5.83),

we generalized factorials to arbitrary real

by using a definition

lim

Nina

(

n-+m

suggested by Euler. Suppose a is a large number; then

lncx!

Ji+mJalnn+lnn!-fln(a+k)),

k=l

and Euler’s summation formula can be used with f(x) = ln(x+ a) to estimate

this sum:

ln(k+ a) =

F,(a,n)

F,(a,O) +

Rz,,,(a,n)

k=l

ln(x + a)

F,(a,x)

(x+a)ln(x+cx-xf

B2k

k=,

2k(2k

1 )(x +

a)2kp1

’

Bz,n(IxI)

RI,,,(a,n) = --

.2m

(x +

a)2m

’

(Here we have used

(9.67)

with a = 0 and b = n, then added ln(n + a)

lna

to both sides.) If we subtract this approximation for xE=, ln(k + a)

from Stirling’s approximation for Inn!, then add alnn and take the limit as

00,

we get

lna! =

alna-a+lnf+o

B2k

B2m(b>)

+fm-l)a2kP1

k=l

2m (x+

OoZrn

’

because

alnn+nlnn-n+i.lnn-(n+a)

ln(nt-a)+n-i

ln(n+a)

-a and

the other terms not shown ‘here tend to zero. Thus Stirling’s approximation

behaves for generalized factorials (and for the Gamma function

a + 1) = a!)

exactly as for ordinary factorials.

Summation 4: A bell-shaped summand.

Let’s turn now to a

sum

that has quite a different flavor:

(9.92)

. +

e-9/n

e-4/n

l/n

e-l/n

-4/n +

e-9/n

. . .

9.6 FINAL SUMMATIONS 469

This is a doubly infinite sum, whose terms reach their maximum value

e”

= 1

when k = 0. We

Cal:1

because it is a power series involving the quantity

eel/” raised to the

(k)th power, where p(k) is a polynomial of degree 2; such

power series are traditionally called “theta functions!’ If n =

1O1oo,

we have

k2/n

0.99005,

when k =

1049;

ec’

0.36788,

when k = 105’;

e-lOO

10P43,

when k = 105’.

So the summand

stays

very near 1 until k gets up to about

fi,

when it

drops off and stays very near zero. We can guess that

will be proportional

fi.

Here is a graph of eekzin when n = 10:

Larger values of n just stretch the graph horizontally by a factor of

$7.

We can estimate

by letting f(x) = eex2/” and taking a =

-00,

b =

$00 in Euler’s summation formula. (If infinities seem too scary, let a = -A

and b = +B, then take limits as A, B

00.) The integral of f(x) is

if we replace x by

u.fi.

The value of

s,”

eeU2

du is well known, but we’ll

call it C for now and come back to it after we have finished plugging into

Euler’s summation formula.

The next thing we need to know is the sequence of derivatives f’(x),

f"(X), . . . .

and for this purpose it’s convenient to set

f(x)

s(x/Jq

g(x) =

epK2

Then the chain rule of calculus says that

df(x)

Q(y)

---

dy dx

’

y = -If_;

and this is the same as saying that

f'(x) =

g'(x/fi).

By induction we have

fck'(x)

nPk’2g(k1(x/fi).

470 ASYMPTOTICS

For example, we have g’(x) = -2xePXL and g”(x) = (4x2

-2)eX2;

hence

f’(x) =

,,-x2/n

)

(

fi-

f”(X) q =

;(4(+)2

m2)e&.

It’s easier to see what’s going on if we work with the simpler function g(x).

We don’t have to evaluate the derivatives of g(x) exactly, because we’re

only going to be concerned about the limiting values when x =

foe.

And for

this purpose it suffices to notice that every derivative of g(x) is

ex2

times a

polynomial in x:

g(k)(x) =

Pk(X)CX2

)

where

is a polynomial of degree k.

This follows by induction.

The negative exponential

e-”

goes to zero much faster than Pk(x) goes

to infinity, when x

!~o3,

so we have

fy+m)

= f’k’(-co) = 0

for all k 3 0. Therefore all of the terms

vanish, and we are left with the term from J f(x) dx and the remainder:

cfi

o(,(‘-m.)/2)

The 0 estimate here follows since

IB,({ufi})

1 is bounded and the integral

ST,”

lP(u)

lePU2

du exists whenever P is a polynomial. (The constant implied

by this 0 depends on m.)

We have proved that

Cfi

O(n”),

for arbitrarily large M; the

difference between

and

Cfi

is “exponentially small!’ Let us therefore

determine the constant C that plays such a big role in the value of 0,.

One way to determine C is to look the integral up in a table; but we

prefer to know how the value can be derived, so that we can do integrals even

9.6 FINAL SUMMATIONS 471

when they haven’t been tabulated. Elementary calculus suffices to evaluate C

if we are clever enough to look at the double integral

+CC

epxz

-M

+CC

e-y’

dy =

+a2

J J

e-(X'+Yz)

dy.

-0c)

-03

-00

Converting to polar coordinates gives

-“T dr

0 0

12"

EZ-

J J

d0 epu

rr.

(u =

l-2)

So C =

,/%.

The fact that

is the equation of a circle whose

circumference is

27rr

somehow explains why

gets into the act.

Another way to evaluate C is to replace x by

and dx by

itt’/2

dt:

+oO

epx2

epxz

J”

t-‘/+t

-DC,

This integral equals

r(i),

since

taP’ePt dt according to (5.84).

Therefore we have demonstrated that

r(i)

,,/?r.

Our final formula, then, is

eCkzin

&iii+

O(neM) ,

for all fixed M.

(9.93)

The constant in the 0 depends on M; that’s why we say that M is “fixed!’

When n = 2, for example, the infinite sum 02 is equal to 2.506628288;

this is already an excellent approximation to

= 2.506628275, even though

n is quite small. The value of

@loo

agrees with

1Ofi

to 427 decimal places!

Exercise 59 uses

ad.vanced

methods to derive a rapidly convergent series

for

0,;

it turns out

,that

@,/&ii

= 1

-I-

2eCnnz

+ 0(

eC4nnL

) .

(9.94)

Summation

5: The clincher.

Now we will do one last sum, which will turn out to tell us the value

of Stirling’s constant

cr.

This last sum also illustrates many of the other

techniques of this last chapter (and of this whole book), so it will be a fitting

way for us to conclude our explorations of Concrete Mathematics.

472 ASYMPTOTICS

The final task seems almost absurdly easy: We will try to find the asymp-

totic value of

by using Euler’s summation. formula.

This is another case where we already know the answer (right?); but

it’s always interesting to try new methods on old problems, so that we can

compare facts and maybe discover something new.

So we THINK BIG and realize that the main contribution to A, comes

from the middle terms, near k = n. It’s almost always a good idea to choose

notation so that the biggest contribution to a sum occurs near k = 0, because

we can then use the tail-exchange trick to get rid of terms that have large

Ikl.

Therefore we replace k by

+ k:

n2;k

(

(2n)!

(n+k)!(n-k)!’

Things are looking reasonably good, since we know to approximate (n

k)!

when n is large and k is small.

Now we want to carry out the three-step procedure associated with the

tail-exchange trick. Namely, we want to write

(2n)!

(n+k)!(n-k)!

ak.(n)

bk(n)

O(Ck(n))

for k

D,,

so that we can obtain the estimate

tbk(n)

ak(n))

bk(n))

O(Ck(n))

k@D,

W’Jn

kED,

Let us therefore try to estimate

(

:Tk) in the region where

Ikl

is small. We

could use Stirling’s approximation as it appears in Table 438, but it’s easier

to work with the logarithmic equivalent in (9.91):

ln ak(n) = ln(2n)!

ln(n

+ k)!

ln(n

k)!

2nln2n-2n+$ln2n+o+O(nP’)

(n+k) ln(n+k) + n + k

ln(n+k)

o + 0(

(n+k)-‘)

(n-k)

ln(n-k)

+ n

ln(n-k)

o + 0( (n-k))‘) .

(9.95)

We want to convert this to

nice, simple 0 estimate.

The tail-exchange method allows us to work with estimates that are valid

only when k is in the “dominant” set D,. But how should we define D,?

9.6 FINAL SUMMATIONS 473

Actually I’m not

into dominance.

We have to make

small enough that we can make a good estimate; for

example, we had better not let k get near n, or the term O((n

k)-‘)

(9.95) will blow up. Yet

must be large enough that the tail terms (the

terms with k

Dn) are negligibly small compared with the overall sum. Trial

and error is usually necessary to find an appropriate set

D,;

in this problem

the calculations we are about to make will show that it’s wise to define things

as follows:

kED,

Ikl

n”‘+‘.

bg6)

Here

is a small positive constant that we can choose later, after we get to

know the territory. (Our 0 estimates will depend on the value of e.) Equation

(9.95) now reduces to

lnok(n) =

(2n+~)ln2-o-~lnn+O(n~‘)

(n+k+i)

ln(l+k/n)

(n--k+:)

141-k/n).

(9.97)

(We have pulled out the large parts of the logarithms, writing

ln(nfk)

lnnfln(1

&k/n),

and this has made a lot of Inn terms cancel out.)

Now we need to expand the terms ln(l

k/n) asymptotically, until we

have an error term that approaches zero as n -+ 00. We are multiplying

ln(

141

k/n) by (n

i),

so we should expand the logarithm until we reach

o(n-‘),

using the assumption that

Ikl

6 n1/2+E:

l*k

*t-$+O(nP3/2+3C).

(

Multiplication by n

k +

yields

;

O(n-“2+3’)

plus other terms that are absorbed in the 0(n-‘/2+3e). So (9.97) becomes

lnok(n) =

(2n+f)ln2-o-~lnn-k2/n+O(nP”2+3’).

Taking exponentials, we have

ak(n)

kw8)

474 ASYMPTOTICS

This is our approximation, with

22n+l/2

bk(n) =

-k2/n

e”fi

’

ck(n)

22nnp1+3e

e-k2/n.

Notice that k enters bk(n) and ck(n) in a very simple way. We’re in luck,

because we will be summing over k.

The tail-exchange trick tells us that

ok(n) will be approximately

bk(n) if we have done a good job of estimation. Let us therefore evaluate

xbk(n)

I-

e-k*/n

(Another stroke of luck: We get to use the sum

from the previous exam-

ple.) This is encouraging, because we know that the original sum is actually

A, =

‘k”

(

)

= (1

1)2, = 22n.

Therefore it looks as if we will have

e“

as advertised.

But there’s a catch: We still need to prove that our estimates are good

enough. So let’s look first at the error contributed by

C&(n):

L(n)

22nn~1+3ee~k2/n

22nn~1+3c@

O(22”npt+3’).

~kl$n’/2+~

Good; this is asymptotically smaller than the previous sum, if

Next we must check the tails. We have

epk2/n

< exp(--Ln’/2+eJ2/n)

+ e-‘/n + e-21” +

)

k>n’/ZiC

O(ep1L2F)

. O(n),

which is O(nPM) for all M;

Eke,,,

bk(n) is asymptotically negligible. (We

chose the cutoff at

n’/2fe

just so that

eek21n

would be exponentially small

outside of D,. Other choices like n

‘/2

logn would have been good enough

too, and the resulting estimates would have been slightly sharper, but the

formulas would have come out more complicated. We need not make the

strongest possible estimates, since our main goal is to establish the value of

the constant o.) Similarly, the other tail

What an amazing

coincidence.

I’m tired of getting

to the end of long,

hard books and not

even getting a word

of good wishes from

the author.

would

be nice to read a

“thanks for reading

this, hope it comes

in handy,” instead

of just running into

a hard, cold, card-

board cover at the

end of a long, dry

proof You know?

9.6 FINAL SUMMATIONS 475

is bounded by 2n times its largest term, which occurs at the cutoff point k

nl/z+E. This term is known to be approximately

(n), which is exponentially

small compared with A,; and an exponentially small multiplier wipes out the

factor of 2n.

Thus we have

;successfully

applied the tail-exchange trick to prove the

estimate

22n

if 0 <

(9.99)

Thanks for reading

We may choose

and conclude that

this, hope it comes

in handy.

-The authors

0 = tln27r.

QED.

Exercises

Warmups

Prove or disprove: If f 1 (n) 4

(n) and

(n) +

(n), then we have

(n)

f2(n)

(n)

92(n).

2 Which function grows faster:

a n(inn)

or (Inn)n?

n(lnln’nn)

or (Inn)!?

c (n!)!

or I((n

l)!)! (n

l)!“!?

FfH,,

HF,?

What’s wrong with the following argument? “Since n = O(n) and 2n =

O(n) and so on, we have

XL=:=,

kn =

Et=,

O(n) =

O(n2).”

Give an example of a valid equation that has O-notation on the left but

not on the right. (Do not use the trick of multiplying by zero; that’s too

easy.) Hint: Consider taking limits.

Prove or disprove: O(f(n) + g(n)) = f(n) + O(g(n)), if f(n) and g(n)

are positive for all n. (Compare with

(g.zT).)

Multiply

(lnn+y+O(l/n))

(n+O(fi)),

and express your answer

in O-notation.

Estimate xkaO eek/” with absolute error O(n-’ ).

Basics

Give an example of functions f(n) and g(n) such that none of the three

relations f(n) 4 g(n), f(n) + g(n), f(n) x g(n) is valid, although f(n)

and g(n)

both.

increase monotonically to

03.

476

ASYMPTOTIC23

9 Prove (9.22) rigorously by showing that the left side is a subset of the

right side, according to the set-of-functions definition of 0.

Prove or disprove: cos O(x) = 1 +

0(x2)

for all real x.

Prove or disprove:

0(x-t

y)2 =

0(x2)

+ O(y2).

12 Prove that

+ O(np2) = (1 +

f)(l

+ O(nm2)),

asn-koc.

13 Evaluate (n + 2 +

O(n

I))”

with relative error O(n ‘).

14 Prove that (n + a)“+P q =

nn+pea(l +

ol(@

-- tcY)n-’ +

O(n2)).

15 Give an asymptotic form.ula for the “middle” trinomial coefficient (n3cn),

correct to relative error

O(n3).

3 ,

Show that if B(l

-x)

= -B(x) 3 0 for 0 < x <

we have

B(H)

f(x)

3 0

if we assume also that f”(x) 3 0 for a < x

17 Use generating functions to show that

B,(i)

= (2’-“’

l)B,,

for all

m 3 0.

18 Find

(2r)(x with relative error O(np’/4), when a > 0.

Homework exercises

Use a computer to compare the left and right sides of the approximations

in Table 438, when n = 10,

= 0.1, and O(f(n)) =

O(f(z))

= 0.

20 Prove or disprove the following estimates, as n

00:

(&J2)

O(lJ;;12)~

e(‘+oc’/n))z

(0(1/n).

n! = O(((1

l/n]nn)n).

21 Equation (9.48) gives the nth prime with relative error O(logn))2. Im-

prove the relative error to O(logn))3 by starting with another term of

(9.31) in (9.46).

22 Improve (9.54) to O(np3).

23 Push the approximation (9.62) further, getting absolute error O(n--3).

Hint: Let

g,,

c/(n

+ 1) (n + 2) +

h,;

what recurrence does h,, satisfy?

9 EXERCISES 477

Suppose a, =

O(f(n)) and b, =

O(f(n)).

Prove or disprove that the

convolution ~~==,

akb+k

is also

O(f(n)),

in the following cases:

f(n) = n-“:, a > 1.

f(n) = a-n,,

> 1.

Prove

(9.1)

and

(9.2),

with which we opened this chapter.

Equation

(9.~1)

shows how to evaluate In

lo!

with an absolute error <

Therefore if we take exponentials, we get

lo!

with a relative

error that is less than

e1/126000000

< 1

Op8.

(In fact, the approximation

gives 3628799.9714.) If we now round to the nearest integer, knowing that

lo! is an integer, we get an exact result.

Is it always possible to calculate n! in a similar way, if enough terms of

Stirling’s approximation are computed? Estimate the value of m that

gives the best approximation to Inn!, when n is a fixed (large) integer.

Compare the absolute error in this approximation with n! itself.

Use Euler’s summation formula to find the asymptotic value of Hipa) =

,E:=,

ka,

where a is any fixed real number. (Your answer may involve a

constant that you do not know in closed form.)

Exercise 5.13 defines the hyperfactorial function Q,, = 1

122

. . . nn. Find

the asymptotic value of Qn with relative error

O(n’).

(Your answer

may involve a constant that you do not know in closed form.)

Estimate the function 1”’ 2’i2 . . .

n””

as in the previous exercise.

Find the asymptotic value of

k’epkL”’

with absolute error

O(n3),

when

is a fixed nonnegative integer.

Evaluate

tk20

l/(ck+cm)

with absolute error O(C-~~), when c > 1 and

m is a positive iinteger.

Exam problems

32 Evaluate

eHn+Hiz’

with absolute error

O(n-‘).

33 Evaluate

tkao

(;)/n’

with absolute error O(np3).

34 Determine values A through F such that (1 + 1 /n)“Hn is

E(lnn)’

Flnn

An+B(lnn)2+Clnn+D+I+-+O(n-‘).

35 Evaluate

I:=,

1 /kHk with absolute error 0( 1).

36 Evaluate

xF=,

l/(n’

+ k2) with absolute error

O(n5).

37 Evaluate

IF=,

mod k) with absolute error O(nlogn).

38 Evaluate tkaO

(i)

with relative error 0

(n-’

) .

478 ASYMPTOTICS

39 Evaluate xOsk<,,

ln(n

-- k)(lnn)k/k! with absolute error

O(n-‘).

Hint:

Show that the terms for k 3

lOInn

are negligible.

40 Let m be a (fixed) positive integer. Evaluate

(-l)kHp

with abso-

lute error O(1).

41 Evaluate the “Fibonacci factorial”

nt=,

with relative error

O(n-‘)

or better. Your answer may involve a constant whose value you do not

know in closed form.

42 Let

be a constant in the range 0 <

We’ve seen in previous

chapters that there is no general closed form for the sum

tksan

(t).

Show that there is, however, an asymptotic formula

whereH(a)=algi+(l--)lg(&).

Hint: Showthat (kn,)

<e(E)

for 0 < k <

OLTI.

43 Show that

C,,

the number of ways to change n cents (as considered in

Chapter 7) is asymptotically

cn4

+ O(n3) for some constant c. What is

that constant?

44 Prove that

as x

0;).

(Recall the definition

x!/(x

i)!

in (5.88), and the

definition of generalized Stirling numbers in Table 258.)

45 Let a be an irrational number between 0 and 1. Chapter 3 discusses the

quantity D(

01,

n), which measures the maximum discrepancy by which

the fractional parts

{kOL)

for 0 6 k < n deviate from a uniform distribu-

tion. The recurrence

D(a,n) 6 D({PI-‘},

[an])

+ (x-’ + 2

was proved in (3.31); we also have the obvious bounds

0 6

D(cc,n)

6 n.

Prove that

lim,,,

OL,

n)/n

= 0. Hint: Chapter 6 discusses continued

fractions.

46 Show that the Bell number

= ee’ &>O

kn/k!

of exercise 7.15 is

asymptotically equal to

where m(n)

m(n) = n

and estimate the relative error in this

approximation.

47 Let m be an integer 3 2. Analyze the two sums

log,

k=l

and

~Ilog,nl

;

which is asymptotically closer to log,,, n! ?

48 Consider a table of the harmonic numbers Hk for 1 < k 6 n in decimal

notation. The kth entry

fi(k

has been correctly rounded to

significant

digits, where

is just large enough to distinguish this value from the

values of

l-lk-1

and

Hk+l

. For example, here is an extract from the table,

showing five entries where

passes 10:

12364 9.99980041-

12365 9.99988128+

12366 9.99996215-

12367 10.00004301-

12368 10.00012386+

Estimate the total number of digits in the table, xc=,

dk,

with an abso-

lute error of 0 (n).

In Chapter 6 we considered the tale of a worm that reaches the end of a

stretching band after n seconds, where H,-1 < 100 6 H,. Prove that if

n is a positive integer such that

H,-l

016

H,, then

Venture capitalists in Silicon Valley are being offered a deal giving them

a chance for an exponential payoff on their investments: For an n mil-

lion dollar investment, where n 3 2, the GKP consortium promises to

pay up to N million dollars after one year, where N =

10n.

Of course

there’s some risk; the actual deal is that GKP pays k million dollars with

probability

(k’H$‘), for each integer k in the range 1 6 k < N. (All

payments are in megabucks, that is, in exact multiples of

$l,OOO,OOO;

the

payoff is determined by a truly random process.) Notice that an investor

always gets at least a million dollars back.

9 EXERCISES 479

480 ASYMPTOTICS

What is the asymptotic expected return after one year, if n million

dollars are invested? (In other words, what is the mean value of the

payment?) Your answer should be correct within an absolute error

I once earned

of O(10-n) dollars.

O(lO-") dollars.

What is the asymptotic probability that you make a profit, if you

invest n million? (In other words, what is the chance that you get

back more than you put in?) Your answer here should be correct

within an absolute error of

O(np3).

Bonus problems

Prove or disprove: j,”

O(xp2)

dx =

O(n’)

as n

co.

Show that there exists a power series A(z) = xk>c

anon,

convergent for

all complex

such that

A(n) +

nn”

Prove that if f(x) is a function whose derivatives satisfy

f’(x)

6 0,

-f"(x)

0, fU'(X)

..)

(-l)"f'"+')(x)

for all x 3 0, then we have

f(x) = f(O)+

yx+...+

:'""~~/xrnel + O(xm),

m .

for x

In particular, the case f(x) =

-ln(l

+ x) proves

(9.64)

for all k,n > 0.

54 Let

f(x)

be a positive, differentiable function such that

xf'(x)

f(x) as

00.

Prove that

f(k)

k3n

k’+‘x

> 0.

Hint: Consider the quantity f(k

l)/(k

i)”

f(k +

i)/(k

i)“.

55 Improve

(9.99)

to relative error

O(n-3/2+5f).

56 The quantity Q(n) =

1 + $i$

+. . . =

nk/nk

occurs in the

analysis of many algorithms. Find its asymptotic value, with absolute

error o(l).

5’7

An asymptotic formula for Golomb’s sum

xka,

/kll

+log,

k]’

is derived

(9.54).

Find an asymptotic formula for the analogous sum without

floor brackets,

xk2,

l/k(

1 +log,

k)2.

Hint: We have j,” uee”ketu du =

l/(1

+tlnk)2.

9 EXERCISES 481

58 Prove that

B,(Ix})

-26

cos(2xk;m-

inrn)

kal

for m 3 2,

by using residue calculus, integrating

-f

2ni

$nize

2ni ---

e2niz

=rn

on the square contour

xfiy,

where max(lxl,

IyI)

M+i,

and letting

the integer M tend to 00.

59 Let

o,(t)

mik+t)‘/n,

a periodic function of t. Show that the

expansion of O,,(t) as a Fourier series is

o,(t)

= &Ei(l + 2eCnJn(cos27rt) + 2e 4XLn(cos4xt)

+2e~9X’n(cos6xt)+...).

(This formula gives a rapidly convergent series for the sum

(0)

in equation (g.g3).)

60 Explain why the coefficients in the asymptotic expansion

(?)

-&(I-&+j&+&-j&g++

all have denominators that are powers of 2.

61 Exercise 45 proves that the discrepancy D(

01,

n) is o(n) for all irrational

numbers

01.

Exhibit an irrational

such that D

(01,

n) is not 0

(n’

’

) for

any

> 0.

Given n, let

{

,;‘,)}

ma& {E} be the largest entry in row n of Stirling's

subset triangle. Show that for all sufficiently large n, we have m(n) =

1777(n)]

or m(n) =

[m(n)],

where

m(n)(%n)

+ 2) In@(n) + 2) = n(K(n) + 1)

Hint: This is difficult.

63 Prove that S.W. Golomb’s self-describing sequence of exercise 2.36 sat-

isfies f(n) =

@2m+n@m1

O(n+-‘/logn).

64 Find a proof of the identity

cos 2n7tx

x7-=

7T2

(x2 -x+ ;) for 0 6 x 6 1,

that uses only “Eulerian” (eighteenth-century) mathematics.

482 ASYMPTOTIC3

Research problems

Find a “combinatorial” proof of Stirling’s approximation. (Note that nn

is the number of mappings of

{

1,2, . . . ,

n} into itself, and n! is the number

of mappings of {1

,2,.

. . , n} onto itself.)

Consider an n x n array of dots, n 3 3, in which each dot has four

neighbors. (At the edges we

“wrap around” modulo n.) Let

be the

number of ways to assign the colors red, white, and blue to these dots in

such a way that no neighboring dots have the same color. (Thus x3 = 12.)

Prove that

Let Q,, be the least integer m such that

> n. Find the smallest

integer n such that Q,, #

[enmy

;I,

or prove that no such n exist.

Th-t/Ah-that’s

all,

folks!

Answers to Exercises

(The first tinder of

every error in this

book will receive

a reward of $2.56.)

Does that mean

I have to find every

error?

(We meant to say

“any error.“)

Does that mean

only one person gets

a reward?

(Hmmm. Try it and

see.)

The number of

intersection points

turns out to give

the who/e story;

convexity was a red

herring.

EVERY EXERCISE is answered here (at least briefly), and some of these

answers go beyond what was asked. Readers will learn best if they make a

serious attempt to find their own answers BEFORE PEEKING at this appendix.

The authors will be interested to learn of any solutions (or partial

solutions) to the research problems, or of any simpler (or more correct) ways

to solve the non-research ones.

1.1

The proof is fine except when n = 2. If all sets of two horses have

horses of the same ‘color, the statement is true for any number of horses.

1.2

If X, is the number of moves, we have

= 0 and X, = X, 1 + 1 +

X,-l

+ 1 +

X,-l

when n > 0. It follows (for example by adding 1 to both

sides) that X, = 3”

1. (After

ix,,

moves, it turns out that the entire tower

will be on the middle peg, halfway home!)

1.3

There are 3c possible arrangements, since each disk can be on any of

the pegs. We must ‘hit them all, since the shortest solution takes 3”

1 moves.

(This construction

:LS

equivalent to a “ternary Gray code,” which runs through

all numbers from (0. . .

0)3

to (2. .2)3, changing only one digit at a time.)

No. If the largest disk doesn’t have to move, 2”

’

1 moves will suffice

ti:

induction); otherwise

(2nP’

1) + 1 +

12nP’

1) will suffice (again by

induction).

1.5

No; different circles can intersect in at most two points, so the fourth

circle can increase the number of regions to at most 14. However, it is possible

to do the job with ovals:

483

484 ANSWERS TO EXERCISES

Venn

[294]

claimed that there is no way to do the five-set case with ellipses,

but a five-set construction with ellipses was found by Griinbaum

[137].

1.6

If the nth line intersects the previous lines in k > 0 distinct points, we

This answer as-

get k- 1 new bounded regions (assuming that none of the previous lines were

SumeS

that

mutually parallel) and two new infinite regions. Hence the maximum number

of bounded regions is

(n-2)+(n-3)+..

. =

S-1

(n-l)(n-2)/2

L,-2n.

1.7

The basis is unproved; and in fact, H(1) # 2.

1.8

B)/a;

= (1

LX+

B)/ct/3;

= (1 + x)/B;

cx;

46 =

So the sequence is periodic!

1.9

(a) We get P(n

1) from the inequality

. . . xn-1

(

+ . . . +

X,-l

n-l

>-(

. .

$X,-l

n-l

(b)

. . .x,x,+1 . ..xz,, 6

(((XI

+ ... + xnl/n)((xn+l + ... + x2nl/n))n by

P(n); the product inside is 6

((x1

+...+xzn)/2n)’

by P(2). (c) For example,

P(5) follows from P(6) from P(3) from P(4) from P(2).

1.10 First show that R, = R,-l + 1 +

Q+l

+ 1 +

R+l,

when n > 0.

Incidentally, the methods of Chapter 7 will tell us that

Q,,

= ((1 +

v’?)~+’

fi)““)/(2fi)

1.11

(a) We cannot do better than to move a double (n

1)-tower, then

move (and invert the order of) the two largest disks, then move the double

1)-tower again; hence A, = 2Anpl + 2 and A,, =

2T,,

2n+1

2. This

solution interchanges the two largest disks but returns the other 2n

2 to

their original order.

(b) Let

be the minimum number of moves. Then

= 3, and it can

be shown that no strategy does better than

= A,-1 + 2 +

A,_1

+ 2 + B,-1

when n >

Hence

2n+2

-5, for all n > 0. Curiously this is just

2A,-1,

and we also have

= A,-1 + 1 + A,-1 + 1 + A,-1 + 1 +

A,-l.

1.12

Ifallmk

>O, thenA(mi

,...,

m,)

=2A(ml,.,,,

m,-1)$-m,. Thisis

an equation of the “generalized Josephus” type, with solution

(ml

, , .

m,,)2

2n-1 ml + . .

.+2m,-1

+m,.

Incidentally, the corresponding generalization of exercise llb appears

to satisfy the recurrence

A(ml,

md,

if m, = 1;

B(ml,...,m,)

2m,-1,

ifn=l;

2A(ml,.

. .

m,-1)

2mn

+B(m,...,m,-l),

ifn>l

andm,>l.

A ANSWERS TO EXERCISES 485

1.13 Given n straight lines that define

regions, we can replace them

by extremely narrow zig-zags with segments sufficiently long that there are

nine intersections between each pair of zig-zags. This shows that ZZ, =

ZZ, ’

+9n-8,

for

a’11

n > 0; consequently ZZ, = 9S,

-8n+

1 = ;n2

In+

1.14 The number

Iof

new 3-dimensional regions defined by each new cut is

the number of 2-dimensional regions defined in the new plane by its intersec-

tions with the previous planes. Hence

’ +

~1,

and it turns out

that

= 26. (Six cuts in a cubical piece of cheese can make 27 cubelets, or

up to

= 42 cuts of weirder shapes.)

Incidentally, the solution to this recurrence fits into a nice pattern if

we express it in terms of binomial coefficients (see Chapter 5):

(;)i-(1;);

L,,

(;)i-(;)-(1);

(‘3)+-(;)-(1)+(Y)

I bet

Iknowwhat

happensin four

Here X, is the maximum number of l-dimensional regions definable by n

dimensions!

points on a line.

1.15

The function I satisfies the same recurrence as J when n >

but I( 1)

is undefined. Since I(2) = 2 and I( 3)

1, there’s no value of I ( 1) =

that

will allow us to use our general method; the “end game” of unfolding depends

on the two leading bits in n’s binary representation.

If n =

2”

2mp1

+k,whereO~k<2m+‘+2m-(2m+2m

‘)=

2”’

+2”’ ‘, the solution is I(n) =

2k+

1 for all n > 2. Another way to express

this, in terms of the representation n =

2”

+ 1, is to say that

I(n) =

{

J(n)

ifO<1<2’“-‘;

J(n)

lrn,

if2”

’

<1<2”.

1.16 Let g(n) =

a(n)ol+

b(n)J3o + c(n)J3’ + d(n)y. We know from (1.18)

that

a(n)x

Wn)h

c(n)61

(1x&,,

, fib,,,

(31,~

i&)3

when

’ . . b’

bo)2;

this defines a(n), b(n), and c(n). Setting g(n) = n in

the recurrence implies that a(n) + c(n)

d(n) = n; hence we know every-

thing. [Setting g(n) = 1 gives the additional identity

a(n)-2b(n)-2c(n)

= 1,

which can be used

tlo

define b(n) in terms of the simpler functions a(n) and

a(n) + c(n).]

1.17

In general we have

W,,,

< 2W,,,

Tk,

for 0 < k < m. (This relation

corresponds to transferring the top n

k, then using only three pegs to

486 ANSWERS TO EXERCISES

move the bottom k, then finishing with the top n

k.) The stated relation

turns out to be based on the unique value of k that minimizes the right-

hand side of this general inequality, when

= n(n +

1)/2.

(However, we

cannot conclude that equality holds; many other strategies for transferring

the tower are conceivable.) If we set

Y,,

(W,,(n+ll,Z

1)/2n,

we find that

Y,,

Y,-1

hence W,(,+1),2

2n(n-

1) + 1.

1.18

It suffices to show that both of the lines from (n2j,0) intersect both of

the lines from (n

0) and that all these intersection points are distinct.

, ,

A line from

(xi,

0) through

(xi

oj, 1) intersects a line from

(xk,

through (Xk

ok,

1) at the

point

(Xj

toj,t)

where t =

(xk

Xj)/(CIk

oj).

Let

n2j

and

= nj + (0 or nP”). Then the ratio t =

(nZk

n2j)/

(nk

(

-nPn

or 0 or

nPn

))

lies strictly between

nj+nk-1

and

nj+nk+l;

hence the y coordinate of the intersection point uniquely identifies j and k.

Also the four intersections that have the same j and k are distinct.

1.19 Not when n >

11.

A bent line whose half-lines run at angles

and

+ 30” from its apex can intersect four times with another whose half-lines

run at angles 4 and

+ 30” only if

> 30”. We can’t choose more

than 11 angles this far apart from each other. (Is it possible to choose

ll?)

1.20

Let h(n) = a(n)o1+ b(n)l& +

c(n)01

+ d(n)yc + e(n)y,. We know

from

(1.18)

that

a(nb+b(n)Bo+c(n)Bl

(aPb,,-,

fib,,-2

. . . obl pbo)4 when

n = (1

b,,-l

. .

bo)z;

this defines a(n), b(n), and c(n). Setting h(n) = n in

the recurrence implies that

a(n)+c(n)-2d(n)-2e(n)

= n; setting h(n) =

implies that a(n) + c(n) + 4e(n) =

n2.

Hence d(n) = (3a(n) +

3c(n)

2n)/4; e(n) = (

a(n)

c(n))/4.

1.21

We can let m be the least (or any) common multiple of

2n,

. * *

)

n + 1. [A non-rigorous argument suggests that a “random” value of m

will succeed with probability

n n-l

&ii

z---&-i..“‘-

n+l

)

-7)

so we might expect to find such an m less than

4n.]

1.22 Take a regular polygon with

sides and label the sides with the

Ioncerode a

elements of a “de Bruijn cycle” of length 2”. (This is a cyclic sequence of

Brudl]

,cyc’e

O’s and l’s in which all n-tuples of adjacent elements are different; see

[173,

exercise

2.3.4.2-231

and

[174,

exercise

3.2.2-171.)

Attach a very thin convex

~Sh~~~~~&~~e,,

’

The Netherlands).

extension to each side that’s labeled 1. The n sets are copies of the resulting

polygon, rotated by the length of k sides for k = 0,

. . . , n

1.23 Yes. (We need principles of elementary number theory from Chap-

ter 4.) Let L(n) = lcm(l,2,. , .

, n). We can assume that n > 2; hence by

A ANSWERS TO EXERCISES 487

Bertrand’s postulate there is a prime p between n/2 and n. We can also

assume that j >

n/2,

since q

’

= L(n) + 1

q leaves j

’

= n + 1

j if and

only if q leaves j. Choose q so that q E 1 (mod L(n)/p) and q

j + 1

(mod p). The people are now removed in order 1, 2, . . . , n

p, j + 1, j + 2,

. . . ,

n,n-p+l, . . . . j-l.

1.24 The only known examples are: X, = a/Xnplr which has period 2;

R. C. Lyness’s recurrence of period 5 in exercise 8; H. Todd’s recurrence

X, = (1 + X, 1 + X,-2)/Xnp3, which has period 8; and recurrences derived

from these by subst.itutions of the form

Y,,

ax,,,,.

An exhaustive search

by Bill Gosper turned up no nontrivial solutions of period 4 when k = 2.

A partial theory has been developed by Lyness

[210]

and by Kurshan and

Gopinath

[189].

An interesting example of another type, with period 9 when

the starting values are real, is the recurrence X, = /X,-l I

X+2

discovered

by Morton Brown

[38].

Nonlinear recurrences having any desired period 3 5

can be based on continuants

[55].

1.25 If Tckl(n) denotes the minimum number of moves needed to transfer n

disks with k auxiliary pegs (hence

T('

)

(n) =

and

TIZi

(n) = Wn), we have

T'ki((n;'))

2T'ki((;))+T'kP'1((k:,)).

No examples (n, k) are known where

this inequality fails to be an equality. When k is small compared with n, the

formula 2”+‘mmk(;:;) gives a convenient (but non-optimum) upper bound on

T’k’((;)).

1.26 The execution-order permutation can be computed in O(n log n) steps

for all m and n

[175,

exercises 5.1.1-2 and 5.1.1-51. Bjorn Poonen

[241]

has

proved that non-Josephus sets with exactly four “bad guys” exist whenever

n E 0 (mod 3) and n 3 9; in fact, the number of such sets is at least

e(i)

for some

> 0. He also found by extensive computations that the only other

n < 24 with non-Josephus sets is n = 20, which has 236 such sets with k = 14

and two with k =

13’.

(One of the latter is

{1,2,3,4,5,6,7,8,11,14,15,16,17};

the other is its reflection with respect to 21.) There is a unique non-Josephus

set with n = 15 and k = 9, namely

{3,4,5,6,8,10,11,12,13}.

2.1

There’s no agreement about this; three answers are defensible: (1) We

can say that

EL=,,

is always equivalent to trnGkc,,

ok;

then the stated

sum is zero. (2) A person might say that the given sum is q4+q3+ql+ql

+qo,

by summing over decreasing values of k. But this conflicts with the generally

accepted convention that

Et=,

= 0 when n = 0. (3) We can say that

x:z,,,

tk+,

tk<,,,

qk;

then the stated sum is

-41

43.

This

convention may appear strange, but it obeys the useful law

xi,,

Et=,+,

EL=,

for all a, b, c.

It’s best to

u;se

the notation

xi=,

only when n

m 3 -1; then both

conventions (1) and (3) agree.

488 ANSWERS TO EXERCISES

2.2 This is

lx].

Incidentally, the quantity ([x > 0]

[x <

01)

is often called

sign(x) or Signum(x); it is +1 when x > 0, 0 when x = 0, and -1 when x < 0.

2.3 The first sum is, of course,

+ al +

as;

the second is

a4+al+ao+al+a4,becausethesumisoverthevalueskEj-2,-1,0,+1,+2}.

The commutative law doesn’t hold here because the function p(k) =

is not

a permutation. Some values of n (e.g., n = 3) have no k such that p(k) = n;

others (e.g., n = 4) have two such k.

2.4

(a)

E:=I

E;=i+J

Et=j+l

aijk

IE:f=j

xfzi+,

~“,++I

aijk

((al23

a1241

a134)

a234.

(b)

x”,==,

x:i,’

x/l;

aijk

x”,=3

x?i

~~~~

aijk

Cl123

-k

(Cl124

(a134

a234)).

2.5

The same index ‘k’ is being used for two different index variables, al-

though k is bound in the inner sum. This is a famous mistake in mathematics

(and computer programming). The result turns out to be correct if

foralljandk,

l<j,k<n.

2.6 It’s

6 j < nl (n

j + 1). The first factor is necessary here because we

should get zero when j < 1 or j > n.

2.7 mx”’ -‘. A version of finite calculus based on V instead of A would

therefore give special prominence to rising factorial powers.

2.8 0, if m 3 1;

l/lm(!,

if m 6 0.

2.9 x”‘+” =

xTfi

(x + m)“, for integers m and n. Setting m = -n tells us

thatx?=l/(x-n)“=l/(x-1)n.

2.10 Another possible right-hand side is Eu Av + v Au.

2.11 Break the left-hand side into two sums, and change k to k + 1 in the

second of these.

2.12 If p(k) = n then n + c = k+

((-l)k

+ 1)c and

((-l)k

+ 1) is even;

hence

(-l)n+c

(-l)k

and k = n-

(-l)“+,c.

Conversely, this value of k

yields p(k) = n.

2.13 Let

(x,

and R, = R,_, +

(-l)“(fi

+ny

+n26)

for n > 0. Then

R(n) = A(n)ol+ B(n)0 +

C(n)y

+ D(n)S. Setting

R,,

= 1 yields A(n) = 1.

Setting R, =

(-1)”

yields A(n)

+2B(n)

(-l)n.

Setting R, =

(-1)“n

yields

-B(n)+ZC(n)

(-1)“n.

Setting R, =

(-l)“n2

yields

B(n)-ZC(n)+

2D(n) =

(-l)“n2.

Therefore 2D(n) =

(-l)“(n2+n);

the stated sum is D(n).

2.14 The suggested rewrite is legitimate since we have k =

~,~i~k

1 when

1 < k < n. Sum first on k; the multiple sum reduces to

(2n+’

2j)

nln+’

(2”+1

l<j<n

“lt is a profoundly

erroneous truism,

repeated by all

copybooks and by

eminent people

when they are

making speeches,

that we should

cultivate the habit

of thinking of what

we are doing. The

precise opposite is

the case. Civiliza-

tion advances by

extending the num-

ber of important

operations which

we can perform

without thinking

about them. Opera-

tions of thought are

like cavalry charges

in a battle-they

are strictly limited

in number, they

require fresh horses,

and must only be

made at decisive

moments.

”

-A. N.

White-

head

/3&Z]

A ANSWERS TO EXERCISES 489

2.15 The first step replaces k(k + 1) by 2

t,5isk

j. The second step gives

GO,,

,, =

(xc=,

k)’

0,.

2.16

x-(x

m)” q =

x2(x

n:im, by (2.52).

2.17 Use induction. for the first two

=‘s,

and (2.52) for the third. The second

line follows from the first.

2.18 Use the facts that (%z)+ 6

lz/,

(%z)) <

/z/,

(?z)+ 6

/zl,

(32))

1~1,

and

Iz/

6 (Rz)+ +

(93~)~

+ (3~)~ + (3~)~.

2.19 Multiply both sides by

2”-l/n!

and let

2”T,/n!=

S,_ 1

+3.2n-’

3(2”

1) +

SO.

The solution is T, =

n! + n!/2nP’. (We’ll see in Chapter 4

that

T,,

is an integer only when n is 0 or a power of 2.)

2.20 The perturbation method gives

Sn +

)&+I

(o&nHkl

+n+l.

2.21 Extracting the final term of &+I gives &+I = 1

S,;

extracting the

first term gives

n+l

(-qn+'

(-y-k

(-l)n+'

(-I)"-k

l<k<n+l

Obkbn

(-l)n+’

Hence 2S, = 1 + (-‘I )” and we have

= In is even]. Similarly, we find

n+l

n+‘I-T,

$(-l)“k(k+l)

T,+S,,

k=O

hence

2Tn

= n + 1 --

and we have T, =

(n + [n is odd]). Finally, the

same approach yields

n+l

(n$-l)‘-U,

Un+2Tn+Sn

+ n + [n is odd] + [n is even]

U,+n+l.

Hence

is the triangular number

(n + 1 )n.

2.22 Twice the sum gives a “vanilla” sum over 1 < j, k 6 n, which splits

into three sums that can be handled easily.

2.23 (a) This approach gives four sums that evaluate to 2n +

2n +

(H,

1). (It would have been easier to replace the summand by

l/k+l/(k+l).)

(b:)

Let u(x)

=2x+1

and Av(x) =

l/x(x+1)

=(x-1)2;

then Au(x) = 2 and v(x) =

-(x

l)-’

-l/x. The answer is

2H,

490 ANSWERS TO EXERCISES

2.24 Summing by parts,

xmHx 6x =

~lnilH,/(m+l)-xm+l/(~+1)2+~;

hence

&k<,,knHk

=nM(H,-l/(m+l))/(m+l)

+Oe/(m+1)2.

our case m = -2, so the sum comes to 1

(H,

l)/(n+

1).

2.25

Here are some of the basic analogies:

na;

&K &K

kc K

~iak+bk!

tbk

kEK

(~akJ(~bk)

okbk

kEK &:K

kCK

ap(kl

apikl

kEK

PiklEK

kEK

PiklEK

~ak[kEK]

nak

,a,EK’

kEK

1 = #K H

nc=c#’

2’26

”

(nl<j,k<n

ajak)

(nl<j=k<n

ojok). The first factor is

(nz=,

a:)2;

the second factor is

nt=,

ai.

Hence P = (nc=, ak)n+'.

2.27

= C~(C

1) =

c*/(c

x). Setting c = -2 and decreasing

x by

yields

A(-(-2)--2)

(-2)“/x,

hence the stated sum is

(-2)L

(-2)n-‘=

(-l)“n!

2.28 The interchange of summation between the second and third lines is

not justifiable; the terms of this sum do not converge absolutely. Everything

else is perfectly correct, except that the result of

tka,

[k = j

k/j should

As opposed to

perhaps have been written

1 >

l](j

1)/j and simplified explicitly.

imperfectly correct.

2.29

Use partial fractions to get

-=-

4k2

+2k-1

The (-l)k factor now makes the two halves of each term cancel with their

neighbors. Hence the answer is

-l/4+

(-1)“/(8n+4).

2.30

tixdx=i(bL-aL)=i(b-a)(b+a~-1).

Sowehave

(b-a)(b+a-1) = 2100 =

22.3.52.7.

A ANSWERS TO EXERCISES

191

There is one solution for each way to write 2100 = x ‘y where x is even and

y is odd; we let a

:1x-yI

and b =

i(x+y)

So the number of

solutions is the number of divisors of 3.

.7, namely 12. In general, there

are

n,,2(n,

+ 1) ways to represent

n,,

pn”, where the products range over

primes.

2.31

tj,k>2j

tja21/j2(l

-l/j)

:=tj,21/j(j-l).

Thesecondsumis,

similarly, 3/4.

2.32

If2n~x~2~n+l,thesumsareO+~~~+n+(x-n-l)+~~~+(x-2n)

n(x-n) = (x-l) + (x-3) +

...

(x-2n+l).

If 2n

1 < x < 2n they are,

similarly, both

equa:i

to n(x

n). (Looking ahead to Chapter 3, the formula

Li(x

l)]

[i(x

l)])

covers both cases.)

2.33 If K is empty,

AkEK

The basic analogies are:

kEK

-t

ak)

kCK

kEK

~!ak+bk!

t%+tbk

min(ak,bk)

ktK

kcK

kEK kEK

min(r\

ak,

bk)

kCK

kEK

kcK

PikICK

apikl

jEJ

kEK

ta,[kEK]

kEK

aplki

PIUEK

ai.k =

aj.k

iEl

ieJ

kEK

Aak.aslkeKi

kEK

permutation that

2.34 Let

K+

= {k 1

3 0} and K = {k 1

0).

Then if, for example, n

consumes terms of

one sign faster than

is odd, we choose

to be

-I U

E,,

where

is sufficiently large that

those of the other

kc(F,

,nK*

)

t-k@,,

tpak)

can steer the sum

toward any value

that it l&s.

2.35 Goldbach’s sum can be shown to equal

as follows: By unsumming a geometric series, it equals xkEP,La, k ‘; there-

fore the proof will be complete if we can find a one-to-one correspondence

between ordered pairs (m, n) with

m,n

3 2 and ordered pairs (k,

with

P and

where

m”

= k’ when the pairs correspond. If m 4 P we let

(m,n)

H (m”, 1); but if m =

P, we let

(m,n)

H (an,b).

492 ANSWERS TO EXERCISES

2.36 (a) By definition, g(n)

g(n

= f(n). (b) By part (a), g(g(n))

With thisself-

g(g(n-

1)) =

tkf(k)[g(n-l)<k~g(n)]

=n(g(n)

-g(n- 1)) =nf(n).

(cl BY part

(

)

~~~~~~e-

ain,

s(s(s(n)))

s(g(g(n-

1))) is

quence

wouldn’t

tf(k)[g(g(n-l))<k6g(g(n))]

do too well on the

Dating Game.

[i=f(k)][g(g(n-l))<k~,g(g(n))]

[i=f(k)][g(n-l)<j~g(n)]

i,k

~j(g(i)-s(i-l))[gin-l)~i~s(n)]

xif(i)

[g(n-l)<i<s(n)]

nxj

[s(n-l)<i<g(n)].

Colin Mallows observes that the sequence can also be defined by the recurrence

f(1) = 1; f(n+ 1) = 1 + f(n+ 1

--f(f(n)))

, for n 3 0.

2.37 (RLG thinks they probably won’t fit; DEK thinks they probably will;

OP is not committing himself.)

3.1

1lgnJ;

L=n-2m=n-211gnl.

3.2 (a)

lx+

.5J. (b) TX-

.51.

3.3 This is Lmn

{mcx}n/aj = mn

since 0 < {ma} < 1.

3.4 Something where no proof is required, only a lucky guess (I guess)

We have

[nxl

nlxl

lnxj

< n

LxJ

+ 1

1x1

:k?<

nLxJ + 1

n{x} < nx < nx

n{x} + 1, by (3.5(a)), (3.7(a)),

(3.7(d)), and (3.8); and this is equivalent to n{x} < 1, when n is a positive

integer. (Notice that

n[xl

6 1nxJ for all x in this case.)

3.6

lf(x,J

lf(Txl)J.

3.7

[n/ml

+ n mod m.

3.8 If all boxes contain <

[n/ml

objects, then n 6

([n/ml

l)m,

n/m + 1 <

[n/ml,

contradicting (3.5). The other proof is similar.

3.9 We have m/n-l/q = (n mumble m)/qn. The process must terminate,

because 0 6 n mumble m < m. The denominators of the representation are

strictly increasing, hence distinct, because

qn/(n

mumble m) > q.

3.10 [x +

[(2x + 1)/4 is not an integer] is the nearest integer to x, if

{x} #

otherwise it’s the nearest even integer. (See exercise 2.) Thus the

formula gives an “unbiased” way to round.

A ANSWERS TO EXERCISES

3.11 If n is an integer,

< n <

[a] < n < [p]. The number of

integers satisfying a < n < b when a and b are integers is (b

1) (b > a).

We would therefore get the wrong answer if 01=

= integer.

3.12

Subtract

[n/m]

from both sides, by (3.6), getting

[(n

mod

m)/m]

[(n

mod m + m

)/ml.

Both sides are now equal to [n mod m >

01,

since

O<nmodm<m.

A shorter but less direct proof simply observes that the first term in

(3.24) must equal the last term in (3.25).

3.13 If they form a partition, the text’s formula for N(cx, n) implies that

+ l/(3 =

because the coefficients of n in the equation N(ol, n) +

N (6, n) = n must agree if the equation is to hold for large n. Hence

and

are both rational or both irrational. If both are irrational, we do get a

partition, as shown in the text. If both can be written with numerator m, the

value m- 1 occurs in neither spectrum. (However, Golomb

[121]

has observed

that the sets

{

LnoLJ

1n 3 l} and

{

[n(3]

-- 1 1 n 3

always do form a partition,

when

l/a+

l/b

= 1.)

3.14 It’s obvious if ny = 0, otherwise true by (3.21) and (3.6).

3.15 Plug in

[mx]

for n in (3.24):

[mx]

[xl

[x-

$1 +

...

e].

3.16

Theformulanmod3=1+f(( ~-~)w”-(w+~)cu~~)

can be verified

by checking it when 0 < n < 3.

A general formula for n mod m, when m is any positive integer, ap-

pears in exercise 7.25.

3.17

,Yj,,[06k<-ml[l

<j<x+k/m]

[k3m(j -XI]

t,lsj<rxl

~j,,lo<k<mIll

6i6

[xl1

Ek[06k<ml

tj=rxl tk[O<k<m(j -xl] =

m[x]

[m(

1x1

x)1 =

-[-mx]

= LmxJ.

3.18 We have

If j 6

nol

1 6 no:

v, there is no contribution, because (j +

~)a-’

< n.

Hence j = [nm] is the only case that matters, and the value in that case

equals

LnK]

+ ~)a.-‘1

n 6

[vol-

‘1.

3.19

If and only if b is an integer. (If b is an integer, log, x is a continuous,

increasing function that takes integer values only at integer points. If

is not

an integer, the condition fails when x = b.)

3.20 We have

kx[cr< kx <

(31

= x

k[ [K/x] <

LB/x]], which sums

~x(lB/xllB/x+

~~/xl~~lx-11).

494 ANSWERS TO EXERCISES

3.21 If 10” < 2M < lo”+‘, there are exactly n+ 1 such powers of 2, because

there’s exactly one n-digit power of 2 for each n. Therefore the answer is

1 +

LMlog2J.

Note: The number of powers of 2 with leading digit

is more difficult,

when 1> 1; it’s ,YOsncM

([nlog2-log11

Lnlog2-log(l+l)]).

3.22 All terms are the same for n and n-l except the kth, where n =

2kP1

and q is odd; we have S, =

S,-1 + 1 and T, = T,-j + 2kq. Hence S, = n

and

T,,

=n(n+ 1).

3.23

Xn=m

im(m-l)<n<im(m+l)

m2-m+i

2n<m2+m++

m-i<fi<m+t.

3.24 Let

= ~x/(ol+ 1). Then the number of times the nonnegative integer

m occurs in Spec( b) is exactly one more than the number of times it occurs

Spec(Lw).

Why? Because N(p,n) = N(K,n)

+n+

3.25 Continuing the development in the text, if we could find a value of m

such that

< m,

we could violate the stated inequality at n + 1 when

In trying to devise

n = 2m + 1. (Also when n = 3m + 1 and n = 3m + 2.) But the existence of

proof

mathe-

such an m = n’ + 1 requires that

2K~,,1,21

< n’ or 3K~,,,31 6 n’, i.e., that

matical induction,

you may fail for

5~2~

1n’Pl

Kl,t/31

ln’I3J

two opposite rea-

sons. You may fail

Aha. This goes down further and further, implying that

6 0; but

= 1.

because you try to

prove too much;

What we really want to prove is that

is strictly greater than n, for

Your

p(n) is

too

all n > 0. In fact, it’s easy to prove this by induction, although it’s a stronger

heavy a burden.

result than the one we couldn’t prove!

Yet you may also

fail because you try

(This exercise teaches an imDortant lesson. It’s more an exercise about

prove

too

little:

Your P(n) is too

weak a support.

the nature of induction than about properties of the floor function.)

3.26 Induction, using the stronger hypothesis

In general, you

have to balance

(($!J"-'))

the statement of

Diq’ < (q-l) for n 3 0.

your theorem so

that the support is

3.27 If D;’ =

2mb-a,

where b is odd and a is 0 or

then DFlb =

3mb-a.

just enough for the

burden.

3.28 The key observation is that a,, = m2 implies an+zk+l = (m+k)2+m-k

and an+2k+2 = (m + k)2 + 2m, for 0 6 k < m; hence an+zm+l = (2m)‘. The

solution can be written in a nice form discovered by Carl Witty:

P6lya

12381

a,-1 = 2’+

[(VT],

when2’+1<n<2’+‘+l+l.

3.29 D(a’, [an]) is at most the maximum of the right-hand side of

~(a’, Lna],y’) =

oL,n,V)+S-e-[Oor

ll-++[Oor

11.

(

A ANSWERS TO EXERCISES 495

This logic is seri-

ously floored.

3.30

K2"

+ a-2", by induction; and X, is an integer.

3.31 Here’s an “elegant,”

“impressive” proof that gives no clue about how

it was discovered:

lxj + lyj +

1YlJ

1x-t

;12YlJ

lx+

il2Yl +

= [2x+ [ZyJ] =

12x1

12YJ.

But there’s also a simple, graphical proof based on the observation that we

need to consider only the case 0 6

x,y

1. Then the functions look like this

in the plane:

A slightly stronger result is possible, namely

1x1

1Yl

l:c+yJ

12x1

+ PYJ

;

but this is stronger only when {x} =

If we replace (x, y) by (-x,x + y ) in

this identity and apply the reflective law (3.4), we get

1YJ

+ lx+yj +

12x1

1x1

12x+2Y1.

3.32 Let f(x) be the sum in question. Since f(x) = f(-x), we may assume

that x 3 0. The terms are bounded by

as k

--oo

and by x2/2k as

+oo, so the sum exists for all real x.

We have

f(2x)

2tk2k-’

I(x/~~-’

[I2

= 2f(x). Let f(x) = l(x) + r(x)

where L(x) is the sum for k 6 0 and r(x) is the sum for k > 0. Then

l(x+

1) =

l(x), and L(x) 6

l/2

for all x. When 0 6 x <

we have r(x) = x2/2

+x2/4

. . .

=x2

andr(x+

1)=(x-

1)2/2+(x+1)2/4+(x+1)2/8+~~~=~2+1.

Hencefix+l)

=f(x)+l,

whenO<x<

We can now prove by induction that f(x+n) = f(x) +n for all integers

n 3 0, when 0 <

< 1. In particular, f(n) = n. Therefore in general,

f(x) =

2-"'f(2"'x)

2~m12mxJ + 2?f({2"x}). But

f({2mx})

1({2"'x})

r({2mx})

+ 1; so If(x)

--xl

6 )2pm12mxJ

--xl

+2pm.t

2-“.5

for all m.

The inescapable conclusion is that f(x) =

Ix/

for all real x.

3.33 Let

n-i

be the radius of the circle. (a) There are

2n-

1 horizontal

lines and

2n-

1 vertical lines between cells of the board, and the circle crosses

each of these lines twice. Since

is not an integer, the Pythagorean theorem

tells us that the cirl:le doesn’t pass through the corner of any cell. Hence

496 ANSWERS TO EXERCISES

the circle passes through as many cells as there are crossing points, namely

4 = 8r. (The same formula gives the number of cells at the edge of the

board.) (b) f(n, k)

=41-j.

It follows from (a) and (b) that

The task of obtaining more precise estimates of this sum is a famous problem

in number theory, investigated by Gauss and many others; see Dickson

[65,

volume 2, chapter

61.

3.34 (a) Let

= [lgn] . We can add 2”’

n terms to simplify the calcula-

tions at the boundary:

f(n)+(2m-n)m

c[lgk]

xj[j=[lgk]][l<k<2m]

k=l

i,k

~j[2jP’ik<2j][l<j<m]

ZjIj-l

2m(m-l)+l.

j=l

Consequently f(n) = nm

2” + 1.

(b) We have [n/21 =

L(n+l)/2],

and it follows that the solution to the

general recurrence g(n) = a(n) + g( [n/21) + g(

[n/2])

must satisfy Ag(n) =

Aa(n)tAg(Ln/2J).

Inparticular,

whena

=n-1, Af(n) =

l+Af(ln/2J)

is satisfied by the number of bits in the binary representation of n, namely

[lg(n + 1 )I. Now convert from A to

A more direct solution can be based on the identities [lg 2jl = [lg j] + 1

and rlg(2j

l)]

= [lgj] +

[j>l],

for j > 1.

3.35

(n+l)2n!e=A,+(n+l)2+(n+l)+B,,where

)‘n!

(n-l)!

multiple

and B, =

(n + l)‘n! + (n + l)‘n! +

+ 2)!

(n+3)!

“’

2el+-

(

1 1

n+3

(n+3)(n+4)

+“’

1 1

l+-

(

nf3

(n+3)(n+3)

+“’

= (n+l)(n+3)

(n + 2)2

is less than 1. Hence the answer is 2 mod n.

A ANSWERS TO EXERCISES 497

3.36 The sum is

2 L4pm[m= Llgl]]

[l=

[lgk]][l

<k~2~“]

k.1.m

22’4

m[2m~L<2m+‘][2L<k<2L+1][O~m<n]

k,L,m

~4~m[2m~1<2m+‘][O~m<n]

2~“[06m<nl

2(1

-2m~n).

3.37 First consider the case m < n, which breaks into

subcases

based on

whether m <

in;

then show that both sides change in the same way when

m is increased by n.

This is really only a

3.38 At most one

x.k

can be noninteger. Discard all integer xk, and suppose

level 4 problem, in

spite of the way it’s

that n are left. When {x} # 0, the average of {mx} as m

lies between

stated.

and

hence {mxl} -t . . + {mx,}

{mxl + . . . + mx,} cannot have average

value zero when n > 1.

But the argument just given relies on a difficult theorem about uniform

distribution. An elementary proof is possible, sketched here for n = 2: Let

be the point ({mx},{my}). Divide the unit square 0 6

x,y

< 1 into

triangular regions A and B according as x + y < 1 or x + y 3 1. We want to

show that

B for some m, if {x} and {y} are nonzero. If

8, we’re

done. Otherwise there is a disk D of radius

> 0 centered at

such that

A. By Dirichlet’s box principle, the sequence PI, . . ,

must contain

two points with

/Pk

Pj/

and k > j, if N is large enough.

It follows that Pk-j

is within

(1,l)

PI; hence Pk-j 1

3.39 Replace j by

j and add the term j = 0 to the sum, so that exercise

15 can be used for th.e sum on j. The result,

[x/bkl

[x/bk+‘] + b

1 ,

telescopes when sum:med on

498

ANSWERS TO EXERCISES

3.40 Let L2J;;I = 4k +

where -2 <

< 2, and let m = LJ;;]. Then the

following relationships can be proved by induction:

segment r m X

-2

2k-1

m(m+l)-n-k

-1

2k-1

-k

m(m+l)-n+k

2k n-m(m+l)+k --k

2k k

n-m(m+l)-k

if and only if

(2kk1)(2k-1)

(2k-1)(2k)

< n < (2k)(2k)

(2k) (2k) < n < (2k)

(2k+l)

(2k)(2k+l)

(2k+1)(2k+l)

Thus, when k 3 1,

is a segment of length 2k where the path travels west

and y(n) = k;

is a segment of length 2k

2 where the path travels south

and x(n) =

-k;

etc. (a) The desired formula is therefore

y(n) =

(-l)“((n-m(m+l))~[~2fi]

isodd]

[irnl).

(b) On all segments, k =

max(ix(n)~,~y(n)~).

On segments wk and

have x < y and n

x + y = m(m + 1) = (2k)’ -- 2k; on segments

and Nk

we have x 3 y and n

y = m(m + 1) = (2k)2 + 2k. Hence the sign is

(-l)lxini<Ylnii

3.41 Since

l/a

l/@2

= 1, the stated sequences do partition the positive

integers. Since the condition g(n) = f(f(n)) + 1 determines f and g uniquely,

we need only show that [[n+]

+ 1 =

lna2j

for all n > 0. This follows

from exercise 3, with

and n = 1.

3.42 No; an argument like the analysis of the two-spectrum case in the text

and in exercise 13 shows that a tripartition occurs if and only if 1

/OL

+ l/(3 +

l/y-l and

{~}+{~}+{~}

= 1,

for all n > 0. But the average value of

{(n-t

1)/a} is

l/2

is irrational, by

the theorem on uniform distribution. The parameters can’t all be rational,

and if y = m/n the average is 3/2

1/(2n). Hence y must be an integer, but

this doesn’t work either. (There’s also a proof of impossibility that uses only

simple principles, without the theorem on uniform distribution; see [125].)

3.43 One step of unfolding the recurrence for K, gives the minimum of the

four numbers 1 + a+ a.b.KLlnPIPaj,(Cbil, where a and b are each 2 or 3.

(This simplification involves an application of (3.11) to remove floors within

floors, together with the identity x + min(y, z) = min(x + y, x + z). We must

omit terms with negative subscripts; i.e., with n

a < 0.)

Too easy.

A more interesting

(still unsolved)

problem: Restrict

both

and

be < 1 , and ask

when the given

multiset determines

the unordered

pair

ia-,

Bl.

A ANSWERS TO EXERCISES 499

Continuing along such lines now leads to the following interpretation:

K, is the least number > n in the multiset S of all numbers of the form

a’ + a’ a2 + a’ a2a3 + . . . + a’ a2a3 . . . a, ,

where m 3 0 and each ok is 2 or 3. Thus,

S =

{1,3,4,7,9,10,13,15,19,21,22,27,28,31,31,...};

the number 31 is in S “twice” because it has two representations

+ 2 + 4 +

8 + 16 = 1 + 3 + 9 + l8. (Incidentally, Michael F’redman

[108]

has shown that

lim,,,

K,/n

= 1,

ie.,

that S has no enormous gaps.)

3 44 Let diqi =

DF!,mumble(q-l),

so that

DIP’

= (qD:_),

+dp))/(q

and

a$’

]D$‘,/(q

-1)l.

Now DF!, 6 (q

1)n

a;’

< n, and the

results follow. (This is the solution found by Euler

[94],

who determined the

a’s and d’s sequentially without realizing that a single sequence

De’

would

suffice.)

3.45 Let 01> 1 sati,sfy

I/R

= 2m. Then we find 2Y, =

a’”

+ aP2”, and

it follows that

= [a’“/21

3.46 The hint follows from (3.g), since

2n(n+

1) =

[2(n+

:)‘I.

Let

n+B

(fi’

fi’-‘)rn

and n’ + 8’ =

(fi”’

&!‘)m,

where 0 <

8,8’

< 1.

Then 8’ = 20 mod 1 = 28

d, where d is 0 or 1. We want to prove that

n’ =

Lfi(n

)]

;

this equality holds if and only if

0 <

e/(2-JZ)+Jz(i

-d) < 2.

To solve the recurrence, note that Spec( 1 + 1

/fi

) and Spec( 1 +

) partition

the positive integers; hence any positive integer a can be written uniquely in

the form a =

\(&’

fi”)m],

here 1 and m are integers with m odd

and

> 0. It follows that

fi’+”

fi”nP’)mj.

3.47 (a) c =

-i.

(1~)

c is an integer. (c) c = 0. (d) c is arbitrary. See the

answer to exercise 1.2.4-40 in

[173]

for more general results.

3.48 (Solution by Heinrich Rolletschek.) We can replace (a,

(3)

({

(3},

\l3J

) without changing

\na]

+ Ln(3]. Hence the condition a =

{B}

necessary. It is also sufficient: Let m =

]-fi]

be the least element of the given

multiset, and let S be the multiset obtained from the given one by subtracting

mn from the nth smallest element, for all n. If a =

{(3),

consecutive elements

of S differ by either

or 2, hence the multiset i.S = Spec(a) determines

01.

3.49 According to unpublished notes of William A. Veech, it is sufficient to

have a(3,

(3,

and 1 linearly independent over the rationals.

500 ANSWERS TO EXERCISES

3.50 H. S. Wilf observes that the functional equation f(x2

1) =

f(x)’

would

determine f(x) for all x 3

if we knew f(x) on any interval

. .

+ e).

3.51 There are infinitely many ways to partition the positive integers into

three or more generalized spectra with irrational ak; for example,

Spec(2ol; 0) U Spec(4cx;

--oL)

U Spec(4a; -301) U

Spec(

fi;

works. But there’s a precise sense in which all such partitions arise by “ex-

panding” a basic one,

Spec(

o1)

Spec(

p); see

[128].

The only known rational

examples, e.g.,

Spec(7; -3) U

Spec(

-1) U

Spec(

0) ,

are based on parameters like those in the stated conjecture, which is due to

A. S. Praenkel

[103].

3.52 Partial results are discussed in

[77,

pages 30-311.

4.1

1, 2, 4, 6, 16, 12.

“Man made

4.2

Note that m,, + n,, =

min(m,,

np)

max(m,,

np). The recurrence

the integers:

~11

e/se

lcm(m,n) = (

n mod m)) lcm(n mod m, m) is valid but not really

advis-

DieudonnC.”

able for computing lcm’s; the best way known to compute

lcm(m,

n) is to

-R. K. Guy

compute gcd(m,n) first and then to divide mn by the gtd.

4.3 This holds if x is an integer, but n(x) is defined for all real x. The

correct formula,

n(x)

X(x

1) = [

1x1

is prime] ,

is easy to verify.

4.4 Between

and

we’d have a left-right reflected Stern-Brocot tree

with all denominators negated, etc. So the result is all fractions m/n with

n. The condition m’n-mn’ = 1 still holds throughout the construction.

(This is called the Stern-Brocot wreath, because we can conveniently regard

the final

as identical to the first

thereby joining the trees in a cycle at

the top. The Stern-Brocot wreath has interesting applications to computer

graphics because it represents all rational directions in the plane.)

4.5

:) and

(Ly)

;

this holds even when k < 0. (We will find a

general formula for any product of L’s and R's in Chapter 6.)

4.6 a = b. (Chapter 3 defined x mod 0 = x, primarily so that this would

After all, ‘mod y’

be true.)

sort of means

“pre-

tend y is zero.” So if

4.7

We need m mod 10 = 0. m mod 9 = k. and m mod

= 1. But m can’t

it already is, there’s

be both even and odd.

nothing to pretend.

A ANSWERS TO EXERCISES 501

4.8 We want 1 Ox

1 Ox

y (mod

15);

hence 5y

0 (mod 15); hence

0 (mod 3). We must have y = 0 or 3, and x = 0 or 1.

4.9

32k+’

mod 4

= 3, so (3

2k+’

-1)/2

is odd. The stated number is divisible

(3’

1)(2

and (3”

1)/2

(and by other numbers).

4.10

999(1

;)(l

-- A) = 648.

4.11

o(O)

= 1;

o(1)

= -1; o(n) = 0 for n > 1. (Generalized Mobius

functions defined on. arbitrary partially ordered structures have interesting

and important properties, first explored by Weisner

[299]

and developed by

many other people, notably Gian-Carlo Rota

[254].)

4.12

xdim

tkid

P(d/k)

g(k)

tk\,,,

td\(m/k)

CL(d)

g(k) =

&,,,

g(k) X

[m/k=

= s(m),

(4.7) and

(4.9).

4.13 (a)

6 1 for all p; (b) p(n) # 0.

4.14 True when k

0. Use

(4.12),

(4.14),

and (4.15).

4.15 No. For example,

mod 5 =

[2or

31;

mod 11 =

[2,3,7,

lo].

4.16

l/e,

+l/e~+~~~+l/e,=l-l/(e,(e,-l))=l-l/(e,+I

-1).

4.17 We have

mod

= 2; hence gcd(f,, f,) = gcd(2,f,) = 1. (Inci-

dentally, the relation

fof,

. ,

f,-l

+ 2 is very similar to the recurrence

that defines the

Eucl.id

numbers e,.)

4.18 Ifn= qmand q isodd, 2”+1 = (2m+1)(2n~m-2n~2m+~~~-2m+1).

4.19 Let

= 2 and let pn be the smallest prime greater than

2Pnm1.

Then

2Pvl

2Pn-I

, and it follows that we can take b =

lim,,,

Igin)

p,,

where

Igin)

is the function lg iterated n times. The stated numerical value

comes from

= 5,

= 37. It turns out that

237

+ 9, and this gives

the more precise value

1.2516475977905

(but no clue about

ps).

4.20 By Bertrand’s, postulate,

10". Let

K =

10PkZPk

.200300005,.

, .

k>l

Then

10nLK

+ fraction (mod

10Znm

').

4.21 The first sum is n(n), since the summand is (k + 1 is prime). The

inner sum in the second is t,Gk<m [k\m], so it is greater than 1 if and only

if m is composite; again we get n(n). Finally

[{m/n}1

= [ntm], so the third

sum is an application of Wilson’s theorem. To evaluate n(n) by any of these

formulas is, of course, sheer lunacy.

502 ANSWERS TO EXERCISES

4.22 (b,”

l)/(b-1)=

((bm-l)/(b-l))(bmn~m+~~~+l).

[Theonly

prime numbers of the form (1

1)/9 for p

2000 occur when p = 2, 19,

23, 317, 1031.1

4.23

p(2k

+ 1) = 0;

p(2k)

= p(k) + 1, for k 3 1. By induction we can show

that p(n) =

p(n-2”),

if n > 2” and m > p(n). The kth Hanoi move is disk

p(k), if we number the disks 0,

. . . , n

1. This is clear if k is a power of 2.

And if 2” < k <

2m+1,

we have p(k) < m; moves k and k

2”’

correspond in

the sequence that transfers m + 1 disks in

T,,,

+ 1 +

T,,,

steps.

4.24 The digit that contributes

dpm

to n contributes dp”-’ + . . . + d =

d(p”‘-

l)/(p

1) to e,(n!), hence

eP(n!)

(n-v,(n))/(p

1).

4.25

n\\n

= 0 or

np,

for all p. It follows that (a) is true.

But (b) fails, in our favorite example m = 12, n =

18.

(This is a common

fallacy.)

4.26 Yes, since

defines a subtree of the Stern-Brocot tree.

4.27 Extend the shorter string with M’s (since M lies alphabetically be-

tween L and R) until both strings are the same length, then use dictionary

order. For example, the topmost levels of the tree are LL < LM < LR <

MM < RL < RM < RR. (Another solution is to append the infinite string

RL” to both inputs, and to keep comparing until finding L < R.)

4.28 We need to use only the first part of the representation:

RRRLL L L L L L R R R R R R

2 3 4 7 10 13. 16 19

22 25 47

91 113 135

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

36~

431"'

The fraction

appears because it’s a better upper bound than

not because

it’s closer than f . Similarly,

is a better lower bound than

The simplest

upper bounds and the simplest lower bounds all appear, but the next really

good approximation doesn’t occur until just before the string of R’s switches

back to L.

4.29 1 /a. To get 1

-x

from x in binary notation, we interchange 0 and 1; to

get 1 /a from a in Stern-Brocot notation, we interchange L and R. (The finite

cases must also be considered, but they must work since the correspondence

is order preserving.)

4.30 The m integers x

[A, A+m) are different mod m; hence their residues

(x mod ml,. . .

, x mod m,) run through all ml . . .

= m possible values, one

of which must be (al mod ml,. . . , a, mod m,) by the pigeonhole principle.

4.31 A number in radix b notation is divisible by d if and only if the sum

of its digits is divisible by d, whenever b = 1 (mod d). This follows because

(a,. . .

aO)b

=a,bm+...+aobo

am+...+ao.

A ANSWERS TO EXERCISES 503

4.32 The

q(m)

nu:mbers

{

kn mod m 1 k

m and 0 < k < m} are the num-

bers {k 1 k

m and 0 6 k < m} in some order. Multiply them together and

divide

nOskcm,

klm

4.33 Obviously h(1) = 1. If m

n then h(mn) =

td,mn

f(d) g(mn/d) =

tc\m,*\n

f(cd) g((m./c)(n/d)) =

Xc,,

Ed,,,

f(c)

s(m/cl

f(d)

g(nld);

this is

h(m) h(n), since c 1. d for every term in the sum.

4.34 g(m) =

x:d,mf(d)

x.d,mf(m/d)

Eda,

f(m/d)

if f(x) is zero

when x is not an integer.

4.35 The base cases are

I(O,n) = 0; I(m,O) = 1.

When

m,n

> 0, there are two rules, where the first is trivial if m > n and

the second is trivial if m < n:

I(m,n)

I(,m,nmodm)

[n/mJI(nmodm,m);

I(m,n)

= I(,m mod n,n) ,

4.36 A factorization of any of the given quantities into nonunits must have

10nZ

:&3,

but this is impossible mod 10.

4.37 Let a, =

2-“ln(e,

5) and b,

2-“ln(e,

i).

Then

n=[E2”+iJ

a,$lnE<b,.

And

a,-1

< a,, < b, < b,

so we can take E =

lim,,,

eaTI.

In fact, it

turns out that

a product that converges rapidly to (l.26408473530530111)2. But these ob-

servations don’t tell us what

is, unless we can find another expression for E

that doesn’t depend on Euclid numbers.

4.38

an-bn=(am-bm)(an~mbO+an~2mbm+...+anmodmbn~m~nmodm)+

bm[n/m]

canmodm

t,nmodm),

4.39 If

. . . at and

. .

. b, are perfect squares, so is

atbl . . .

b,/cf

. . c: ,

where

{al

, . . .

,at}n{bl,...,b,}={cl,...

,cV}.

(It can be shown, in fact, that

the sequence

(S(l),S(2),S(3),.

. . , )

contains every nonprime positive integer

exactly once.)

504 ANSWERS TO EXERCISES

4.40 Let f(n) =

n,,,,,,,,,

k =

n!/pl"/pJ

Ln/p]! and g(n) = n!/pEP(“!l.

Then

s(n)

f(n)f(

ln/PJ)

ln/p’J)

f(n)

b/d)

Also f(n) = ao!(p

l)!Ln/pl =

ao!(-l)L"/PJ

(mod

p),

and

e,(n!)

Ln/pJ

(Ln/pJ

. These recurrences make it easy to prove the result by induction.

(Several other solutions are possible.)

4.41 (a) If

= -1 (mod p) then (n2)(pP’i/2 = -1; but Fermat says it’s

+l.

(b) Let n = ((p

1)/2)!; we have n = (--l)(P~‘i’2

n,sk<p,2(p

-k) =

l)!/n,

hence

= (p

l)!.

4.42 First we observe that k

+=+ k

ak for any integer a, since

gcd(k,

= gcd(k,

ak) by Euclid’s algorithm. Now

m-Ln

and

n’In

mn’

-L

mn’+nm’

Similarly

m’

n’ and n-!-n’

mn’+ nm’

n’.

Hence

n and m’

-L

n’ and n

n’

mn’+nm’

nn’.

4.43 We want to multiply by LP’R, then by

RP’

LV’RL, then L-’ R, then

RP2LP’RL2, etc.; the nth multiplier is RPpcnlLP’RLp”“, since we must cancel

p(n) R’s. And

Rm~mL

‘RLm =

(y,;:,).

4.44 We can find the simplest rational number that lies in

[.3155,.3165)

= [$$&,a)

by looking at the Stern-Brocot representations of

and $$$ and stopping

just before the former has L where the latter has

(ml,nl,m2,n2)

:= (631,2000,633,2000);

while

n.1

rn2

rnz

then (output(L);

(nl,nz)

(ml,m,))

else (output(R);

(ml,

m2) :=

(ml,

ml)

(nl

,nz))

The output is LLLRRRRR = &

.3158. Incidentally, an average of .334

implies at least 287 at bats.

A ANSWERS TO EXERCISES 505

4.45 x2 E x (mod

10n)

(

x(x

1) E 0 (mod 2”) and x(x

1) E 0

(mod

5n)

x mod 2” = [Oor

and x mod 5” = [Oor 11. (The last step

is justified because x(x

1) mod 5 = 0 implies that either x or x

1 is a

multiple of 5, in which case the other factor is relatively prime to 5n and can

be divided from the congruence.)

So there are

,at

most four solutions, of which two (x = 0 and x = 1)

don’t qualify for the title “n-digit number” unless n = 1. The other two

solutions have the forms x and 1

+ 1 -- x, and at least one of these numbers

is > 1 On-‘. When n = 4 the other solution, 10001

9376 = 625, is not a

four-digit number.

1Ne

expect to get two n-digit solutions for about 90% of

all n, but this conjecture has not been proved.

(Such self-reproducing numbers have been called “automorphic.“)

4.46 (a) If j’j

k’k = gcd(j,k), we have nk’knscdii,k) = ni’i = 1 and

nk’k

1. (b) L te n = pq, where p is the smallest prime divisor of n. If

2” E 1 (mod n) then 2”

1 (mod p). Also 2P-l = 1 (mod p); hence

2scdipm

‘,nl

= 1 (mod p). But gcd(p

1 In) = 1 by the definition of

4.47 If

n+’

= 1 (mod m) we must have n

m. If

for some

1 < j < k < m, then nkPj = 1 because we can divide by nj. Therefore if the

numbers

n’

mod m, . ,

n”-’

mod m are not distinct, there is a k < m

with

= 1. The least such k divides m-

by exercise 46(a). But then kq =

1 )/p for some prime p and some positive integer q; this is impossible,

since nkq $ 1. Therefore the numbers

n’

mod m, . . , nmP’ mod m are

distinct and relatively prime to m. Therefore the numbers

. , m

1 are

relatively prime to

n-L,

and m must be prime.

4.48 By pairing numbers up with their inverses, we can reduce the product

(mod m) to

l~n<m,n2modm=l

Now we can use our knowledge of the

solutions to

mod m =

By residue arithmetic we find that the result is

1 if m = 4, pk, or

2pk

(p > 2); otherwise it’s

+l.

4.49 (a) Either m < n (@(N

1) cases) or m = n (one case) or m > n

(O(N

1) again). H

ence

R(N) = 2@(N

1) + 1. (b) From (4.62) we get

2@(N-l)+l

l+x~(d)LN/d]LN/d-11;

d>l

hence the stated result holds if and only if

p(d)

LN/dJ

= 1 ,

d2’

for N

And this is a special case of (4.61) if we set f(x) = (x 3 1)

506 ANSWERS TO EXERCISES

4.50 (a) If f is any function,

f(k)

f(k)[d=gcd(km)]

@$k<m

d\m

OSk<m

f(k)

[k/d1

m/d]

d\m

OSk<m

f(kd)[kIm/d]

d\m

O<k<m/d

f(km/d)[kI

;

d\m

OSk<d

we saw a special case of this in the derivation of (4.63). An analogous deriva-

tion holds for

instead of

Thus we have

(z-

Wk)

(z-

mkm’d) =

n?&&)

OSk<m

d\m

OSk<d

d\m

k_Ld

because

w”‘/~

e2ni’d.

Part (b) follows from part (a) by the analog of (4.56) for products

instead of sums. Incidentally, this formula shows that

Y,(z)

has integer

coefficients, since

Y,(z)

is obtained by multiplying and dividing polynomials

whose leading coefficient is 1.

4.51

(x~+...+x,)~

tk,+...+k,,zpp!/(kl!.

k,!)x:’

. .

.x:,

andthecoeffi-

cient is divisible by p unless some

= p. Hence

(x1

.+x,)P

E x7

.+xK

(mod p). Now we can set all the x’s to 1, obtaining

E n.

4.52 If p > n there is nothing to prove. Otherwise x

xkcP

‘I

(mod p); this means that at least

[(n

l)/(p

~ l)] of the given numbers are

multiples of p. And (n

l)/(p

1) 3 n/p since n 3

4.53 First show that if m 3 6 and m is not prime then (m-2)!

0 (mod m).

(If m = p2, the product for (m

2)! includes p and

2p;

otherwise it includes

d and m/d where d < m/d.) Next consider cases:

Case 0, n < 5. The condition holds for n = 1 only.

Case 1, n > 5 and n is prime. Then (n

l)!/(n + 1) is an integer and

it can’t be a multiple of n.

Case 2, n 3 5, n is composite, and n + 1 is composite. Then n and

n+l divide

(n-l)!,andnIn+l;

hencen(n+l)\(n-l)!.

Case 3, n > 5, n is composite, and n + 1 is prime. Then (n

l)! E 1

(mod n + 1) by Wilson’s theorem, and

[(n-l)!/(n+l)J

((n-l)!+n)/(ntl);

A ANSWERS TO EXERCISES 507

this is divisible by

11.

Therefore the answer is: Either n = 1 or n # 4 is composite.

4.54

(1 OOO!) > 500 and

(1 OOO!)

249, hence 1 OOO! = a. 1

0249

for some

even integer a. Since 1000 = (1300)5, exercise 40 tells us that

2249

looo!/5249

-1 (mod 5). Also 2

249

= 2, hence a = 2, hence a mod 10 = 2_

or 7; hence the answer is 2.1

0249.

4.55 One way is to prove by induction that

P&Pt(n

+ 1) is an integer;

this stronger result helps the induction go through. Another way is based

on showing that each prime p divides the numerator at least as often as it

divides the denominator. This reduces to proving the inequality

k=l

which follows from

k=l

[(In

)/ml

LWmj

lnhl

The latter is true when 0 6 n < m, and both sides increase by 4 when n is

increased by m.

4.56 Let f(m) = ~~~~’

min(k,2n-k)[m\k],

g(m) =

EL=:

(2n-2k-1)

[m\(2k+

l)].

number of times p divides the numerator of the stated

product is f(p) + f(p2) + f(p3) +

...,

and the number of times p divides the

denominator is g(p) +

g(p2)

g(p3)

...

. But f(m) = g(m) whenever m

is odd, by exercise 2.32. The stated product therefore reduces to 2”‘”

‘1,

exercise 3.22.

4.57 The hint suggests a standard interchange of summation, since

[d\ml

[m=

dkl = Ln/dj .

lSVlI$II

O<k<n/d

Calling the hinted sum ,X(n), we have

I(m

+ n)

X(m) ~ X(n) =

v(d).

dES(m,nl

On the other hand, we know from (4.54) that

,X(n)

in(n

+ 1). Hence

.X(m

+ n)

X(m) ~ Z(n) = mn.

4.58 The function f(m) is multiplicative, and when m =

it equals 1 +

p + +

pk.

This is a power of 2 if and only if p is a Mersenne prime and

k =

For k must be odd, and in that case the sum is

(1 +p)(l

+p2

+p4

+-+pk

‘)

508 ANSWERS TO EXERCISES

and (k- 1)/2 must be odd, etc. The necessary and sufficient condition is that

m be a product of distinct Nersenne primes.

4.59 Proof of the hint: If

= 1 we have

= 2, so there’s no problem.

If n > 1 we can assume that

6 . . < x,. Case 1:

xi’

+ . . . +

xi!,

(x,

1))’

3 1 and x, > x+1. Then we can find

3 x,

1 3

x,-l

such

that

xl’

+ ... +x;l,

+P-'

= 1; hencex, 6

p-t1

andxl...x,

. . .

~~~1

+ 1) 6

. . . e,, by induction. There is a positive integer m

such that a =

. . . x,/m; hence a 6

. . .

= e,+l

1, and we have

. . .

~~(~~+l)<el...e,e,,+l.

Case2:

x~'+~~~+x~~,+(~,-l)-~~l

and

x,,

x,-l.

Let a = x, and a-’ + (a

)-’

= (a

2))’ + L-l. Then

we can show that a 3 4 and (a-2)(<+ 1) 3

a2.

So there’s a

such

that

xi’

+ ...

+ x,1, + (a -- 2)-l + p-’ = 1; it follows by induction that

. . .

..x.-2(a-2)(2+

..x+z(a-2)(@

+ 1)

..e.,

and we can finish as before. Case 3:

XT’

+ . . . +

xi!,

+ (x,

I)-'

Let a =

xn,

and let a-’ +

0~~’

= (a

)-’

(?-‘.

It can be shown that

+ 1) > a( a + 1

because this identity is equivalent to

aa2-a’a+aa-a2+a+a

> 0,

which is a consequence of

aa( a

(1 + a)a

( 1

a)a

Hence

we can replace

and a by a

1 and

(3,

repeating this transformation until

cases 1 or 2 apply.

Another consequence of the hint is that

l/x,

+ . . . + l/x, < 1 implies

l/xl

...

+1/x,

l/e1

...

+1/e,; see

exercise 16.

4.60 The main point is that

Then we can take

sufficiently large

(to meet the conditions below) and

to be the least prime greater than

p;-,.

With this definition let a,, =

33”lnp,

and

3-nln(p,

+ 1). If we

can show that a,-1 < a,, <

6 b,-1, we can take P =

lim,,,

can

as in

exercise 37. But this hypothesis is equivalent to

pi-,

p,,

< (p,-l +

l)3.

there’s no prime

p,,

in this range, there must be a prime p <

p;-,

such that

p +

cpe

> (p,-1 + 1

)3.

But this implies that

cpe

3p213,

which is impossible

when p is sufficiently large.

We can almost certainly take

= 2, since all available evidence indi-

cates that the known bounds on gaps between primes are much weaker than

the truth (see exercise 69). Then

= 11,

1361,

= 2521008887, and

1.306377883863 < P

1.306377883869.

4.61 Let

T?L

and

be the right-hand sides; observe that fin’

m’fi

= 1,

hence

??I.

T?.

Also

m/c

m//n’ and N = ((n + N )/n’)n’

n 3

((n+N)/n’-l)n’-

--n

N n’ 3 0. So we have

T?-L/?L

3 m/‘/n”. If equality

doesn’t hold, we have n” = (&n’

m’fi)n” = n’( tin”

m”fi) + fi(m”n’

m’n”) 3 n’ +

> N, a contradiction.

A ANSWERS TO EXERCISES 509

I have discovered a

wonderful proof of

Fermat’s Last Theo-

rem, but there’s no

room for it here.

Therefore, if Fer-

mat’s Last Theorem

is false, the universe

will not be big

enough to write

down any numbers

that disprove it.

Incidentally, this exercise implies that (m +

m”)/(n

+ n”) =

m//n’,

although the former fraction is not always reduced.

4.62 2 ‘$2

2+2

-2

6-2

7+2~'2+2~~'3-2~20-2~21+2~30+2~31-

~42

+ . . . can be written

;

+ 3

t(2-4k1-6k-3

2-4k2-10k

-7)

k>O

Incidentally, this sum can be expressed in closed form using the “theta func-

tion” O(z,

e~xhkz+2irk; we have

e t-3

~;6(~ln2,3iln2)

&O(%ln2,5iln2)

4.63 Any n > 2 either has a prime divisor d or is divisible by d = 4. In either

case, a solution with exponent n implies a solution (an/*)*+(bn/*)* = (c”/*)*

with exponent d. Since d = 4 has no solutions, d must be prime.

The hint follows from the binomial theorem, since

aP+(x-a)P-pap

’

is a multiple of x when p is odd. Assume that a

-L

x. If x is not divisible

by p, x is relatively prime to cP/x; hence x = mp for some m. If x is divisible

by p, then

cp/x

is divisible by p but not by p2, and

has no other factors

in common with x.

(The values of a, b, c must, in fact, be even higher than this result

indicates! Inkeri

[160]

has proved that

A sketch of his proof appears in

[249,

pages 228-2291, a book that contains

an extensive survey of progress on Fermat’s Last Theorem.)

4.64 Equal fractions in

appear in “organ-pipe order”:

2m 4m rm 3m m

2n’ 4n’ . . . .

. . . . 3n’

Suppose that

IPN

is correct; we want to prove that

&+I

is correct. This

means that if kN is odd, we want to show that

k-l

N+l

yN,kN;

if kN is even, we want to show that

k-l

yN,kN 1 yN,kN

N+l

yN,kN

YN,kN+l

510 ANSWERS TO EXERCISES

In both cases it will be helpful to know the number of fractions that are

strictly less than (k

l)/(N + 1) in

LPN;

this is

i(kN-dtl),

d =

gcd(k-l,N+l),

by (3.32). Furthermore, the number of fractions equal to (k

l)/(N + 1) in

~PN

that should precede it in

iPN+l

even]), by the nature of

organ-pipe order.

IfkNisodd,thendisevenand(k-l)/(N+l)isprecededbyt(kN-l)

elements of

?N;

this is just the correct number to make things work. If

even, than d is odd and (k

)/(

is preceded by

(kN ) elements of

?N.

If d =

none of these

equ’als

l)/(N

+ 1) and

‘J’N,~N

‘<‘;

otherwise

(k- 1

)/(

1) falls between two equal elements and

~PN

,k~

is ‘=‘. (C. S. Peirce

[230]

independently discovered the Stern-Brocot tree at about the same time

as he discovered

?N.)

4.65 The analogous question for the (analogous) Fermat numbers

is a

“N

square less

famous unsolved problem. This one might be easier or harder.

than 25 x

1014

divides a Euclid

4.66 It is known that no square less than 36 x

1018

divides a Mersenne

number.”

number or Fermat number. But there has still been no proof of Schinzel’s

--I/an

Vardi

conjecture that there exist infinitely many squarefree Mersenne numbers. It

is not even known if there are infinitely many p such that

p\\(

a h b), where

all prime factors of a and b are < 31.

4.67 M. Szegedy has proved this conjecture for all large n; see

[284’],

[77,

pp.

78-791,

and

[49].

4.68 This is a much weaker conjecture than the result in the following ex-

ercise.

4.69 Cram&

[56]

showed

t’hat

this conjecture is plausible on probabilistic

grounds, and computational experience bears this out: Brent

[32]

has shown

that

P,+l

< 602 for

Pn+l

< 2.686 x

1012.

But the much weaker bounds

in exercise 60 are the best currently proved

[221].

Exercise 68 has a “yes”

answer if

P,+j-P,

2PA"

for all sufficiently large n. According to Guy

[139,

problem

A8],

Paul Erdas

offe:rs

$10,000 for proof that there are infinitely many

n such that

clnn

lnlnn lnlnlnlnn

n+l

-P,>

___

(lnlnlnn)2

A ANSWERS TO EXERCISES 511

for all c > 0.

4.70 This holds if and only if

(n) = 1/3(n), according to exercise 24. The

methods of

[78]

may help to crack this conjecture.

4.71 When k = 3 the smallest solution is n = 4700063497 = 19.47.5263229;

no other solutions are known in this case.

4.72 This is known to be true for infinitely many values of a, including -1

(of course) and 0 (not so obviously). Lehmer

[199]

has a famous conjecture

that cp(n)\(n

1) if and only if n is prime.

4.73 This is known to be equivalent to the Riemann hypothesis (that all

zeros of the complex zeta function with real part between 0 and 1 have real

part equal to

l/2).

What’s

114

radix 11

4.74 Experimental evidence suggests that there are about

1 /e) dis-

tinct values, just as if the factorials were randomly distributed modulo p.

5.1 (11): =

(14641),,

in any number system of radix

3 7, because of the

binomial theorem.

5.2 The ratio

(Karl)/

= (n-

k)/(k+

1) is < 1 when k 3 Ln/2J and 3 1

when k <

[n/2],

so the maximum occurs when k = [n/2] and k =

[n/2].

5.3

Expand into factorials. Both products are equal to f(n)/f(n

k)f(k),

where f(n) =

(n+

l)!n!

(n- l)!.

5.4

(-,‘) =

(-l)k(k+;P’)

(-l)‘(;)

(-l)k[k>O].

If 0 < k < p, there’s a p in the numerator of (E) with nothing to cancel

the denominator. Since (E) =

(“i’)

(:I;),

we must have

(“i’)

(-l)k

(mod p), for 0 < k

5.6

The crucial step (after second down) should be

The original derivation forgot to include this extra term, which is [n =

01.

512 ANSWERS TO EXERCISES

5.7

Yes, because

(--1 )

"/(

-r

1)".

We also have

rqr

+ i)" =

(2r)9;!%

5.8 f(k) = (k/n

1)” is a polynomial of degree n whose leading coefficient

nn.

(5.40)~

the sum is n!/nn. When n is large, Stirling’s approxima-

tion says that this is approximately

&/en.

(This is quite different from

l/e), which is what we get if we use the approximation (1

-k/n)”

eek,

valid for fixed k as n

oo.)

5.9

E,(z)t

ksO

t(tk + t)k-‘zk/k! =

tk.Jk

l)k

'(tz)k/k!

(tz),

(5.60).

5’1o tk>O

2zk/(k + 2) =

F(2,l;

3; z), since tk+l/tk = (k +

2)z/(k

+ 3).

5.11 The first is Besselian and the second is Gaussian:

But not

Imbesselian.

z-~‘sinz =

tka,(-l)kz2k/(2k+1)!

= F(l;l,i;-z2/4);

’

arcsin

tkZo z2k(;)k/(2k+

l)k! =

F(;,

;;

5;~~).

5.12 (a) Yes, the term ratio is n.

(b) No, the value should be 1 when

k = 0; but (k +

1)"

works, if n is an integer. (c) Yes, the term ratio is

(k+l)(k+3)/(k+2).

(d) No, the term ratio is 1

+l/(k+l)Hk;

and

Ink

isn’t a rational function. (e) Yes, the term ratio is

t(k+ 1)

T(n

t(k)

T(n

.I)

’

(f) Not always; e.g., not when t(k) =

and T(k) = 1. (g) Yes, the term ratio

can be written

at(k+l)/t(k)

bt(k-t2)/t(k)

ct(k+3)/t(k)

a+bt(k+l)/t(k)

+ct(k+2)/t(k)

’

and t(k+m)/t(k) = (t(k+m)/t(k+m-1)) . . . (t(k+

1)/t(k))

is arational

function of k.

5.13 R, = n!n+‘/Pi = Qn/P,, =

Qi/n!“+‘.

5.14 The first factor in (5.25) is

(,‘i

k,)

when k < 1, and this is (-1

)Lpkpm

(r-r::).

The sum for k 6

is the sum over all k, since m 3

(The condition

n 3 0 isn’t really needed, although k must assume negative values if n < 0.)

To go from (5.25) to (5.26), first replace s by -1 -n

5.15 If n is odd, the sum is zero, since we can replace k by n-k. If n =

2m,

the sum is

(-1)“(3m)!/m!3,

by (5.29) with a = b = c = m.

A ANSWERS TO EXERCISES 513

5.16 This is just (;!a)! (2b)!

(2c)!/(a+

b)!

(b+c)!

(c+ a)! times (5.2g), if we

write the summands in terms of factorials.

5.17 (

27/2)

(;;)/22”;

(2-/2)

(;;)/24”;

(2nn’/2)

22n(2-/2).

5.18

(:;)(,"k",k)i33k.

5.19

.t(-2)

’

tkzO

(kP’,“P’)

(-l/(k

~ tk

1)) (-.z)~, by (5.60), and

this is tkaO

(tt)(l/(tk-

1))~~

‘BH,(z).

5.20 It equals

F(-al,

. ,-a,;

-bl,

. . .

,-b,;

(-l)mfnz);

see exercise 2.17.

5.21

lim,,,(n

m)c/nm

5.22 Multiplying and dividing instances of (5.83) gives

(-l/2)!

x!(x-l/2)!

("n'")

(n+xc1'2)n

2r/(n-i'2)

= lim

(

2n + 2x

n--2X

n+cc

’

by (5.34) and (5.36). Also

1/(2x)! = lim

2n+2x

(

)

(2n)

-2x

n--c%

Hence, etc. The Gamma function equivalent, incidentally, is

T(x)

l-(x

= r(2x) r(;)/22x-

’

5.23 (-l)"ni, see (5.50).

5.24 This sum is (1:) F(

",~~"ll>

(fz),

(5.35)

and

(5.93).

5.25

This is equivalent to the easily proved identity

a’

(a+llEemb<

(a-b)-- =

(b +

II)”

(b+l)k

as well as to the operator formula a

b = (4 + a)

(4 + b).

Similarly, we have

(al

a21 F

al,a2,a3,

bl,

= alF

al+l,a2,a3,...,am

-wF(

al,a2+l,a3,...,am

bl,

. , b,

514 ANSWERS TO EXERCISES

because

(a1

+ k)

(al + k). If al

is a nonnegative integer d,

this second identity allows us to express F(al , , . , a,,,;

, . . . ,

b,;

z) as a lin-

ear combination of F(

+ j, a3, . . .

, a,,,; b2, . ,

b,;

z) for 0 6 j 6 d, thereby

eliminating an upper parameter and a lower parameter. Thus, for example,

we get closed forms for F( a, b; a

1; z), F(

a, b; a

2; z), etc.

Gauss

[116,

$71

derived analogous relations between F(a, b; c;z) and

any two “contiguous” hypergeometrics in which a parameter has been changed

. Rainville [242’] gene:ralized this to cases with more parameters.

5.26 If the term ratio in the original hypergeometric series is

tkfl

/tk = r(k),

the term

ratio

in the new one is tk+I/tk+l =

r(k

+ 1). Hence

(

al,

. . . ,

a, al+l,...,a,+l,l

bl,

. . . .

"-'amzF

..b.

(

,...!

b,+l,2

’

5.27 This is the sum of the even terms of

F(2a1,.

. .

,2a,;

2bl,.

. .

,2b,;

z).

We have

(2a)=

/(2a)X

4(k+

a)(k+ a +

i),

etc.

5.28

WehaveF(“;b]z)=

(I-z)~"F(~~~~"~~)=

(l-zzpuF('

,"sals)=

mz)c

bF(C-yb

1~).

(Euler proved the identity by showing that both

sides satisfy the same differential equation. The reflection law is often at-

tributed to Euler, but it does not seem to appear in his published papers.)

5.29 The coefficients of 2” are equal, by Vandermonde’s convolution. (Kum-

mer’s original proof was different: He considered

lim,,,

F(m, b

a; b; z/m)

in the reflection law

(5.101).)

5.30 Differentiate again to get

z(1

z)F"(z)

3z)F'(z)

F(z) = 0.

Therefore

F(z)

F(l,l;2;z)

'by (5.108).

5.31 The condition f(k) = cT(k+ 1)

CT(k) implies that f(k+

1)/f(k)

(T(k+2)/T(k+

l)/(l

--T(k)/T(k+

1)) is a rational function of k.

5.32 When summing a polynomial in k, Gosper’s method reduces to the

“method of undetermined coefficients!’ We have q(k) = r(k) = 1, and we

try to solve p(k) =

s(k+

s(k). The method suggests letting s(k) be a

polynomial whose degree is cl = deg(p) + 1.

5.33 The solution to k = (k- l)s(k+ 1)

(k+ l)s(k) is s(k) = -k+ 5;

hence the answer is

2k)/‘2k(k

1) + C.

5.34 The limiting relation holds because all terms for k > c vanish, and

c cancels with

-c

in the limit of the other terms. Therefore the second

partial sum is

lim,,o

F(-m,--n;

e-mm;l)

lim,,~(e+n-m)m/(e-m)m

(-l)yy).

5.35 (a)

2m"3n[n>0].

(b)

i)PkP’[k>,O]

=2k+‘[k>0].

A ANSWERS TO EXERCISES 515

The boxed

sentence

on the

other side

of this page

is true.

5.36 The sum of the digits of m + n is the sum of the digits of m plus the

sum of the digits of n, minus p

1 times the number of carries, because each

carry decreases the digit sum by p ~ 1.

5.37 Dividing the first identity by n! yields (‘ly) =

(i) (,yk),

Van-

dermonde’s convolution. The second identity follows, for example, from the

formula

(-l)k(-~)”

if we negate both x and

5.38 Choose c as large as possible such that

(5)

< n. Then 0 < n

(5)

(‘3

(3;

replace n by n

(i)

and continue in the same fashion.

Conversely, any such representation is obtained in this way. (We can do the

same thing with

n =

(9’)

(“;-)

f...+

(z),

0 6

.”

< a,

for any fixed m.)

5.39 xmyn =

~~=,

(“‘+l-:

k)anbm~kxk

I;=,

(m+~~~~~k)an

kbmyk

for

all mn > 0, by induction on m + n.

5.40 (-l)m+’

x;=,

I;,

(;)(”

= (--l)m+’

I;=,((”

“k;i

S-l)-

(-y))

(-qm+l((rn

i-1

)

("--'y1))

('2")

(L),

5.41

tkaOn!/(n

k)! (n + k +

l)!

(n!/(2n

l)!)

tk>,,

(‘“k+‘),

which is

2’%!/(2n

+ 1

)!.

5.42 We treat n as an indeterminate real variable. Gosper’s method with

q(k) = k + 1 and r(k) = k

1 -n has the solution s(k) =

l/(n

+ 2); hence

the desired indefinite sum is (-1

)XP’

$$/(“z’).

And

This exercise, incidentally, implies the formula

n-l

(

)

(n+ll’(kyl)

(n+;)(L)

’

a “dual” to the basic recurrence (5.8).

5.43 After the hinted first step we can apply (5.21) and sum on k. Then

(5.21) applies again and Vandermonde’s convolution finishes the job. (A com-

binatorial proof of this identity has been given by Andrews [lo]. There’s a

quick way to go from this identity to a proof of (5.2g), explained in [173,

exercise 1.2.6-621.)

516 ANSWERS TO EXERCISES

5.44 Cancellation of factorials shows that

(;)(;)(“;“)

(m+;:;-k)(j;k)(y;;).

so the second sum is

l/(

“:“I

I times the first. We can show that the first sum

is (“ib) (“-~P~~b),

whenever n 3 b, even if m < a: Let a and b be fixed

and call the first sum S( m, ‘1). Identity (5.32) covers the case n = b, and

we have S(m,n) =

S(m,n

1) + S(m- 1,n) +

(-l)m+n(mzn)(i)(“)

since

(m+;:;-k)

r+:;J;j--k)

(m-;‘;r;-k).

The result follows by induction on

m+ n, since

(,)

= 0 when n > b and the case m = 0 is trivial. By symmetry,

the formula (“ib) (m+l:im

“)

holds whenever m > a, even if n < b.

5.45 According to (5.g),

xc<,,

(kPi’2) =

(n+A’2).

If this form isn’t “closed”

enough, we can apply (5.35) and get (2n + 1) (‘,“)4-“.

5.46 By (5.6g), this convolution is the negative of the coefficient of

z2*

%‘(z)K’(-z).

Now (223-‘(z)

1)(2%‘(-2)

1) =

dm;

hence

‘K’(z)‘E_

i-z)

ad-7

i’%‘(z)

iZ3

-1(-z)

$. By the binomial

theorem,

16~~)"~ =

(-16)“z2”

-t

(;)

so the answer is (2~)4”~‘/(2r~

1) +

(4;;1)/(4n

1).

5.47 It’s the coefficient of z” in (IBr(~)“/Qr(~))(~Br(~)~s/Qr(~)) =

l/Q,(z)‘,

where Qr(z) = 1

-r

rBB,(z'i

by (5.61).

5.48 F(2n + 2,1; n + 2;

22n+‘/(2z:;), a special case of (5.111).

5.49 Saalschiitz’s identity (5.97) yields

5.50 The left-hand side is

k+a+m-1

and the coefficient of 2” is

A ANSWERS TO EXERCISES 517

by Vandermonde’s convolution (5.92).

5.51 (a) Reflection gives F(a,

-n;

2a; 2) = (-1 )“F( a,

-n;

2a; 2). (Inciden-

tally, this formula implies the remarkable identity

A2”‘+’

f(0) = 0, when

f(n) = 2nxc/(2x)“.>~

(b) The term-by-term limit is

&kSm

(r)

m(-2)k

plus an addi-

tional term for k = 2m

1: the additional term is

(-m)... (-1) (1)...(m)

(-2m+

1) . . . (-1)22m+’

I:-2m).

(-1) (2m

l)!

pm+1

(-ltm+'*

-2

(CL')

hence, by (5.104), this limit is

-l/(

y2),

the negative of what we had.

5.52 The terms of both series are zero for k > N. This identity corresponds

to replacing k by N

k. Notice that

5.53 When b =

-i,

the left side of (5.110) is 1

22 and the right side is

-42+422)"2, independent of a. The right side is the formal power series

l/2

(

)

42(2-l)+

l/2

(

16z2(z-1)2+~~~,

which can be expanded and rearranged to give 1

22+

Oz2

Oz3

;

but the

rearrangement involves divergent series in its intermediate steps when

so it is not legitimate.

5.54 If m + n is odd, say 2N

we want to show that

lim F

E'O

(

N-m-;,

-N+c

-m+e

=o.

Equation (5.92) applies, since -m +

> -m

i +

and the denominator

factor

T(c-b)

T(N-m)

is infinite since N < m; the other factors are finite.

Otherwise m + n is even; setting n = m ~ 2N we have

fi,mo

(

-N,

N-m-i+e

(N-1/21N

-m+c

rnN

by (5.93). The remaining job is to show that

l/2)!

(m-N)!

-(-l/2)!

m! =

518 ANSWERS TO EXERCISES

and this is the case x = N of exercise 22.

5.55 Let Q(k) =

(k+Al)...(k+AM)Zand

R(k) =

(k+Bl)...(k+BN).

Then t(k+ 1)/t(k) =

P(k)Q(k-

l)/P(k-

l)R(k),

where P(k) = Q(k) -R(k)

is a nonzero polynomial.

5.56 The solution to

-(k+l)(k+2)

s(k+l)+s(k)

is s(k) =

-ik2-k-a;

hence

(~:)

6k=

i(-l)kp’(2k2

+4k+

1) + C. Also

(-l)k-’

k-t

‘+‘r”*>

(,+,-

‘-)*)

v(2k2+4k+1)+;

5.57 We have

t&+1)/t(k)

(k-n)(k+l

+B)(-z)/(k+l)(k+O).

Therefore

we let p(k) = k+ 8, q(k) = (k-

n)(-z),

r(k) = k. The secret function s(k)

must be a constant

0~0,

and we have

k+B

(-z(k-n)--k)as;

hence

010

-l/(1

+ z) and

-nz/(l

+ z). The sum is

(;)zk(“-+6k

-&(;~;)z’+c

(The special case z = 1 was mentioned in (5.18); the general case is equivalent

(5.1311.)

5.58 If m > 0 we can replace

(:)

(;I\)

and derive the formula

T,,,

$T,,-I,~-~

(“i’).

The summation factor

(t)-’

is therefore appropriate:

We can unfold this to get

m,n

To,n-m

H,-,

Finally

To,~

,,,

H,.

,,,,

T,,,,

(z)

(H,

-H,).

(It’s also possible to derive

this result by using generating functions; see Example

in Section 7.5.)

5.59

)*O,kal

(y)[j=Llognrkj] =

ti>0,k>,

(~)[m’<k<mj+‘l, which is

tj>O

(‘j’)(mj+’

mj) =

(m--

l)tjao

(3)mj

= (m-

l)(m+

l)n.

A ANSWERS TO EXERCISES

519

5.60 (‘c)

4n/&K is the case m = n of

(my)

/gq(l

;)n(l

G)?

5.61 Let [n/p] = q and n mod p = r. The polynomial identity (x + 1

xp + 1 (mod p) implies that

(x+

pq+r

(~+l)~(x~

+l)q

(mod p).

The coefficient of

x”’

on the left is (E). On the right it’s

(,I,,)

(z), which

is just

(

mbd

(

,m~tpJ)

because 0 6

5.62 (,‘$) = ,&i~,,.+k,,=mp

(kg)

. . .

(zn)

(E)

(mod p’), because all terms

of the sum are multiples of

except the

(i)

terms in which exactly m of the

k’s are equal to p. ((Stanley

[275,

exercise 1.6(d)] shows that the congruence

actually holds modulo

when p > 3.)

5.63 This is

~~=,(-4)k(~+~)

~~=,(-4)nPk(2n~k).

The denomina-

tor of (5.74) is zero when

-l/4,

so we can’t simply plug into that formula.

The recurrence

-2&-l

-.SnP2 leads to the solution

(-l)n(2n+l).

5.64

~,,,((;k) +

(2;+,))/@+

(;$,)/(k+

11,

which

A.&

(g;:)

= '",';;"

5.65 Multiply both sides by nn-’ and replace k by n

k to get

(7Y)

n-1

nk(n

k)! = (n

l)!

Z(nkf’/k!

nk/(k

l.)!)

k=O

(n-l)!nn/(n-l)!.

(The partial sums can, in fact, be found by Gosper’s algorithm.) Alternatively,

knnPlekk! can be interpreted as the number of mappings of

, . . . , n} into

itself with

(1))

. . . ,

f(k)distinctbutf(k+l) l {f(l),...,f(k)};summingonk

must give nn.

5.66 This is a

“wa.lk

the garden path” problem where there’s only one “ob-

vious” way to proceed at every step. First replace k

j by

then replace

[A]

by k, getting

j&o

(j:‘k)

(A)

’

I ,

520 ANSWERS TO EXERCISES

The infinite series converges because the terms for fixed j are dominated by

a polynomial in j divided by 2j. Now sum over k, getting

Absorb the j + 1 and apply (5.57) to get the answer,

4(m+

1).

5.67

3(2nntt52)

by (5.26), because

(‘i’)

3(y).

5.68 Using the fact that

we get

“(2”

’

(,zij,)).

[n is even] ,

5.69 Since

(k:‘)

+ (‘y’) < (:) + (i)

k < 1, the minimum occurs

when the k’s are as equal as possible. Hence, by the equipartition formula of

Chapter 3, the minimum is

(n mod m) -t (n

(n mod m))

b/ml

-4

mod m)

A similar result holds for any lower index in place of 2.

5.70 This is F(-n,

1;2); but it’s also

(-2)Pn(F)F(-n,

-n;

if we

replacekbyn-k.

NowF(-n,-n;i-n;:)

=F(-f,-l;&n;l)byGauss’s

identity (5.111). (Alternatively, F(-n,-n;

i-n;

2-“F(-n,

i-n;

-1)

by the reflection law (5.101), and Kummer’s formula (5.94) relates this to

(5.55).) The answer is 0 when n is odd, 2-“(,,y2) when n is even. (See

[134,

$1.21

for another derivation. This sum arises in the study of a simple search

algorithm [

1641.)

5.71 (a) S(z) = EkZO okzm-+k/(l -Z)m+Zk+’ = Zm(l

-2)

-“-‘A@/(1

-z)‘).

(b) Here A(z) =

k20

(2,“)(-z)k/(k + 1) =

(dm

1)/2z,

so we have

A(z/(l

-z)‘)

= 1

-z.

Thus

[z”]

(z/(1

2))“’ =

(;I;).

5.72 The stated quantity is m(m

n) . . . (m

l)n)nkPYik’/k!. Any

prime divisor p of n divides the numerator at least k

y(k) times and di-

vides the denominator at most k

v(k) times, since this is the number of

A ANSWERS TO EXERCISES 521

times 2 divides k!.

prime p that does not divide n must divide the prod-

n at eas as often as it divides k!, because

uc;-tL;)-n)...(m-(k-l)

)

1 t

..(m-(p’-1)

)’

a multiple of

p’

for all

3 1 and all m.

5.73 Plugging in X, = n! yields

= 1; plugging in X, = ni yields

= 0. Therefore the general solution is X, =

olni

+ b(n!

ni).

5.74

(“l’)

(;I:),

for 1 6 k 6 n.

5.75 Therecurrenc:e

Sk(n+l)

Sk(n)+S

)

mod

(n) makes it possible to

verify inductively

th’at

two of the S’s are equal and that

.S-,I

mod3(n) differs

from them by (-1)“. These three values split their sum So(n) +

(n) +

.Sz(n)

= 2n as equally as possible, so there must be 2” mod 3 occurrences of

[2”/31 and 3

(2” mod 3) occurrences of 12”/3J.

5.76

Qn,k

l(c) +

(kn+,)’

5.77 The terms are zero unless kl 6 .. <

k,,

when the product is the

multinomial coefficient

(

kl,

. . . ,

k,pl

’

Therefore the sum over kl , . . . ,

k,-l

mkm

, and the final sum over

yields

(

mn+’

l)/(m- 1).

5.78 Extend the sum to k = 2m2 + m

1; the new terms are

(1)

(‘,)

...-t

(1;‘)

= 0. Since m

(2m+

l),

the pairs (kmod m,kmod

(2m-t

1))

are distinct. Furthermore, the numbers (2j + 1) mod

(2m+

1) as j varies from

0 to 2m are the numbers 0,

. . . ,

2m in some order. Hence the sum is

5.79 (a) The sum is 22np’, so the gcd must be a power of 2. If n = 2kq where

q is odd, (:“) is divisible by

2k+’

and not by

2k+2.

Each

(:$)

is divisible

by 2k+’ (see exercise 36), so this must be the gtd. (b) If

p’

6 n + 1 < p’+‘,

we get the most radix p carries by adding k to n

k when k =

p’

1. The

number of carries in this case is

e,(n

l),

and

e,(L(n

+ 1)).

5.80 First prove by induction that k! 3 (k/e)k.

5.81 Let fL,m,n(x) be the left-hand side. It is sufficient to show that we have

fl,,,,(l)

> 0 and tlhat

f;,,,,(x)

< 0 for 0 < x 6 1. The value of

fl,,,,(l)

(-l)"p"p'(':~~")

(5.23),

and this is positive because the binomial

coefficient has exactly n

m- 1 negative factors. The inequality is true when

= 0, for the same reason. If

> 0, we have

f&,+(x)

-Iftpl,m,n+l(~),

which is negative by induction.

522 ANSWERS TO EXERCISES

5.82 Let ~,,(a) be the exponent by which the prime p divides a, and let

m = n

k. The identity to be proved reduces to

For brevity let’s write this as

min(x,,yl,zl)

min(xz,y2,z2).

Notice that

= x2 +

22.

The general relation

+(a)

e,(b)

e,,(a) =

eP(/u*bl)

allows us to conclude that x.1 # x2

==+

min(x,

,x2) = 0; the same holds also

for

(~1,

y.7)

and (2, ,22). It’s now a simple matter to complete the proof.

5.83 If m < n, the quantity (j:“)

(“‘?:iPk)

is a polynomial in k of degree

less than n, for each fixed

.i;

hence the sum over k is zero. If m 3 n and

is an integer in the range n <

6 m, the quantity

(‘+kk)

(m+c:iPk) is a

polynomial in j of degree less than r, for each fixed k; hence the sum over j is

zero. If m 3 n and if

= -d

1 is an integer, for 0 6 d < n, we have

(;)

(w(qd)

(-lIq;)(;);

hence the given sum can be written

pk(i;k)(;)(:)(;I)(m+;lI:-k)

pk(;)

(‘:“)

(jy)

(m+;:;

-“>

&,,+m+-l

j,kL

(k)(;)(‘:“)(-‘i”l’)(-“m’*,‘)

xc-1

)k+mi-L

k,i

(;)

(7”)

(-‘mn12).

This is zero since (I:“) is a polynomial in k of degree d < n.

If m 3 n, we have verified the identity for m different values of r. We

need consider only one more case to prove it in general. Let

= 0; then j = 0

and the sum is

pk(;)

(-+;;-

“) =

by (5.25). (Is there a substantially shorter proof?)

A ANSWERS TO EXERCISES

523

5.84 Following the hint, we get

andasimilarformulafor&,(z).

Thustheformulas

(ztB;‘(z)‘B[(z)+l)Bt(z)r

and

(ztE;‘(z)&:(z)

+ l)&,(z)’ give the respective right-hand sides of (5.61).

We must therefore prove that

(zwwJm

l)%w-

_ t

:‘%

(z)

(zw4~:M

+ 1)Wz)’ = ,

&Z)t

and these follow from (5.59).

5.85 If f(x) =

a,x”

...

+ a’x +

is any polynomial of degree < n, we

can prove inductively that

c-1

“+“‘+‘“f(e1x,

+...+E,x,)

(-l)nn!~,I~l

..x..

O$f,

,...,

E$,$l

The stated identity is the special case where a, = 1 /n! and

= k3.

5.86 (a) First expand with n(n- 1) index variables

Lij

for all i # j. Setting

kii = li’

-Lji

for 1

i < j < n and using the constraints

tifi

(lij

-iii)

= 0 for

all i < n allows us to carry out the sums on

li,

for 1 6 j < n and then on

iii

for 1 < i < j < n by Vandermonde’s convolution. (b) f(z)

1 is a polynomial

of degree < n that has n roots, so it must be zero. (c) Consider the constant

terms in

,jJsn

;)“’

JIn

;)“‘~

ifi

i#i

5.87 The first term is

(n;k)zmk, by (5.61). The summands in the second

term are

k20

(n+

1)/m;

(l+l/m)k)iiz),.;,

= --

(‘+‘;~~~;‘-‘)(i,jk.

524 ANSWERS TO EXERCISES

Since ~06j<m(<‘i+‘)k =

m(--l)‘[k=mL],

these terms sum to

(l+l/mW

-n-

(-z”)k

mk-n--l

k>n/m

)

(m+l)k-n-

k>n/m

)

(-zm)k

(”

-kmk)pk

k>n/m

Incidentally, the functions ‘B,,,(zm) and L2i+‘z~B1+II,(L2~+‘~)‘~m are the m+l

complex roots of the equation w”‘+’

wm = z”l.

5.88 Use the facts that

Jr(e

nt)

dt/t

= Inn and (1

ee’)/t

$ 1.

(We have

(“,)

= O(kmx ‘)

a,s

00, by (5.83); so this bound implies that

Stirling’s series

(i)

converges when

> -1. Hermite

[155]

showed that

the sum is In r( 1 + x).)

5.89 Adding this to (5.19) gives

~~‘(x+y)~+’

on both sides, by the binomial

theorem. Differentiation gives

sentence

the

other side

of this page

is not

self-

referentia/.

and we can replace k by k + m + 1 and apply (5.15) to get

(m;::

(-‘I;

‘)

(-X)m+l+ky--l-k-n

In hypergeometric form, this reduces to

which is the special case (a, b, c, z) = (n +

m + 1 +

m + 2, -x/y) of the

reflection law (5.101). (Thus (5.105) is related to reflection and to the formula

in exercise 52.)

5.90 If

is a nonnegative integer, the sum is finite, and the derivation in

the text is valid as long as

:none

of the terms of the sum for 0 < k <

has

zero in the denominator. Otherwise the sum is infinite, and the kth term

ml-

‘)

(

-i-l)

is approximately

k”

’

(-s

l)!/(-r

l)!

by (5.83). So we

A ANSWERS TO EXERCISES 525

need r > s+ 1 if the infinite series is going to converge. (If r and s are complex,

the condition is

’31s

+ 1, because

lkZl

= km’.) The sum is

-r, 1

(

I)-

r(r-s-l)T(-s)

s+l

-S

T(r-s)T(-s-l)

s+l-r

by (5.92); this is the same formula we found when r and s were integers.

5.91 (It’s best to use a program like MACSYMA for this.) Incidentally,

when c = (a+

1)/2,

this reduces to an identity that’s equivalent to (5.110), in

view of the Pfaff’s reflection law. For if w =

-z/(

-2)

we have 4w( 1

w) =

-42/(1

z)‘, and

ia,

a+;-b

1 +a-b

4w(l-w)

(l-z)uF(,;;~bir).

5.92 The identities can be proved, as Clausen proved them more than 150

years ago, by showing that both sides satisfy the same differential equation.

One way to write the resulting equations between coefficients of z” is in terms

of binomial coefficients:

(I;)

L’k)

Ck)

r+skl/2)(r+s11/2)

n k

Another way is in terms of hypergeometrics:

a,b,

i-a-b-~--n

(Za)“(a+b)“(2b)“;

i+a+b,l-a-n,l-b-n

(2a+2b)“a”b”

$+a,

tb,a+b-n,-n

l+a+b,i+a-n,i+b-n

(1/2)“(1/2+a-b)“(l/2-a+b)”

+a+b)n(l/4-a)K(1/4-b)”

’

5.93

0~~

’

n:_,

(f(j) +

cx)/f(j).

(The special case when f is a polynomial of

degree 2 is equivalent to identity (5.133).)

526 ANSWERS TO EXERCISJES

5.94 This is a consequence of Henrici’s “friendly monster” identity,

f(a,z)f(a,wz)f(a,w"z)

(

;a-+,

ia++

5a,3a+~i,~a+5,3a-~,fa,~a+~,a

where f (a, z) =

F(;

a; z). This identity can be proved by showing that both

sides satisfy the same differential equation. If we replace 3n by 3n + 1 or

3n + 2, the given sum is zero.

5.95 See

[78]

for partial

reisults.

The computer experiments were done by

A. Vyssotsky.

5.96 All large n have the property, according to S&k&y [256’]. Paul ErdGs

conjectures that, in fact, ma+,

((2,“)) tends to infinity as n

00.

5.97 The congruence surely holds if 2n + 1 is prime. Steven Skiena has also

Ilan Vardi notes

found the example n = 2953, when 2n + 1 =

3.11

.179.

that the condi-

tion holds for

6.1

2314,2431,3241,1342,3124,4132,4213,1423,2143, 3412,4321.

2n+l

=p’,

where p is prime,

6.2

{

E}n-&,

because every such function partitions its domain into k

non-

if and only if

empty subsets, and there are

rnk

ways to assign function values for each

2pm’

mod

= 1.

partition. (Summing over k gives a combinatorial proof of (6.10).)

This yields two

more examples:

6.3 Now

dk+’

6 (center of gravity) --E = 1

-e+(d’

+...+dk)/k.

This

(‘093~-‘)/2;

recurrence is like (6.55) but with 1

in place of 1; hence the optimum

n =

(35112-1)/2.

solution is

dk+’

= (1

c)Hk. This is unbounded as long as

< 1.

6.4 Hln+’

:H,,.

(Similarly

EC”=,

(-l)kp’/k

= Hz,,

H,.)

6.5

U,(x,y)

is equal to

ky)n-‘.

The

k31

(;)(-l)kp'(~+ky)n-'

This proves (6.75). Let

R,(x,y)

x~“U,(x,y);

then

Ro(x,y)

0 and

R,(x,y)

R,-'(x,y)

l/n+y/x,

hence

R,(x,y)

H,+ny/x.

(Incidentally, the original sum

U,(n,

-1) doesn’t

lead to a recurrence such as this; therefore the more general sum, which de-

taches x from its

dependenice

on n, is easier to solve inductively than its

special case. This is another instructive example where a strong induction

hypothesis makes the difference between success and failure.)

The Fibonacci

re-

currence is additive,

6.6 Each pair of babies bb present at the end of a month becomes a pair

but

the

rabbits

are

of adults

at the end of the next month; and each pair

becomes an

multiplying.

A ANSWERS TO EXERCISES 527

If the harmonic

and a

bb.

Thus each bb behaves like a drone in the bee tree and each

behaves like a queen, except that the bee tree goes backward in time while

the rabbits are going forward. There are F,+l pairs of rabbits after n months;

of them are adults and

F,-,

are babies. (This is the context in which

Fibonacci originally introduced his numbers.)

numbers are

worm

numbers, the Fi-

6.7 (a) Set k = 1 -- n and apply (6.107). (b) Set m = 1 and k = n- 1 and

bonacci numbers

are rabbit numbers.

apply (6.128).

6.8 55 + 8 + 2 becomes 89 + 13 + 3 = 105; the true value is 104.607361.

6.9 21. (We go from

to F,+z when the units are squared. The true

answer is about 20.72.)

6.10 The partial quotients

a~,

al, az, . . . are all equal to 1, because

1 + 1

/c$.

(The Stern-Brocot representation is therefore RLRLRLRLRL.. .

6.11

(-1)” = [n=O]

[n=l];

see (6.11).

6.12

This is a consequence of (6.31) and its dual in Table 250

6.13 The two formulas are equivalent, by exercise 12. We can use induction.

Or we can observe that znDn applied to f(z) =

gives

xnzX

while 9” applied

to the same function gives xnzX; therefore the sequence (a’, 4’

,a2,.

. . ) must

relate to (z”Do,z’D’, z2D2,. . . ) as (x0,

x1,x2,.

) relates to (x”, x1,

x2,.

.).

6.14 We have

x(“i”)

(k+l)(~~~)

+In-k)(x~~~l),

because (n+l)x= (k+l)(x+k-n)+(n-k)(x+k+l). (It suffices toverify

the latter identity when k = 0, k = -1, and k = n.)

6.15 Since A((‘Ak)) =

(iTi),

we have the general formula

= A”(x”) =

(-l)mpi(x

+ j)”

Set x = 0 and appeal to (6.19).

6.16

An,k

tj>o

{

“i’};

this sum is always finite.

6.17

(a) [;I = [l:T.!,]. (b) /:I =

= n!

[n3

k]/k!. (c)

IL/

k!(z).

6.18 This is equivalent to (6.3) or (6.8). (It follows in particular that

o,(l) =

-na,(O)

= U&n! when n > 1.)

6.19 Use Table 258.

6’20

xl<j<k<n

l/j2 =

t,4jsn(n+

j)/j’

= (n + l)Hp)

H,.

528 ANSWERS TO EXERCISES

6.21 The hinted number is a sum of fractions with odd denominators, so

it has the form a/b with a and b odd. (Incidentally, Bertrand’s postulate

implies that b, is also divisible by at least one odd prime, whenever n > 2.)

6.22

Iz/k(k

+ z)I <

2/21/k;’

en k >

21~1,

so the sum is well defined when

the denominators are not zero. If z = n we have I:=:=, (l/k

l/(k

+ n)) =

Hm+n

H,,

which approaches

as m -3

co.

(The quantity HZ-r

is often called the psi function Q(z).)

6.23

z/(e’+l)

=z/(e’-

I)-2z/(e2’-1)

=tRaO(l

-2n)B,zn/n!.

6.24 When n is odd, T,,(x) is a polynomial in x2, hence its coefficients

are multiplied by even num.bers when we form the derivative and compute

T,+l (x) by (6.95). (In fact we can prove more: The Bernoulli number

B2,,

always has 2 to the first power in its denominator, by exercise 54; hence

22n

\\Tl,,+r

2k\\(n

i- 1). The odd positive integers (n + 1

)TJ,+~

/22n

are called Genocchi numbers (1,

1,3,17,155,2073,.

. . ), after Genocchi [117].)

6.25

lOOn-nH,

< lOO(n- 1)

(n-

l)H,-1

H,-l

> 99. (The least

such n is approximately

e99m~Y,

while he finishes at N =

eloom

about

times

as long. So he is getting closer during the final 63% of his journey.)

6.26 Let u(k) =

HkP1

and Av(k) = l/k, so that u(k) = v(k). Then we have

Hi2’

I;=,

Hkp,/k

Hip, I;+’

HL,

5,.

6.27 Observe that when

T~I

> n we have

gcd(F,,F,)

= gcd(F,

,,F,)

(6.108). This yields a proof by induction.

6.28 (a) Q,, = ol(L, ~ F,,)/2 +

fiFn.

(The solution can also be written

Q,, =

cxF,

1 + BF,.) (b)

+”

$“.

6.29 When k = 0 the identity is (6.133). When k =

it is, essentially,

K(xI,.

x,)x, = K(x,, . . .

,x,)

K(x,,

. . .

,x,)

K(x,, . . . ,

~m~~)K(xm+z,...,xn);

in Morse code terms, the second product on the right subtracts out the cases

where the first product has intersecting dashes. When k > 1, an induction

on k suffices, using both (6.127) and (6.132). (The identity is also true when

one or more of the subscripts on K become -1, if we adopt the convention that

K 1 = 0. When multiplication is not commutative, Euler’s identity remains

valid if we write it in the form

K,+.,(xl,...,x,+,iKk(X,+k,...,Xm+l)

Km+k(Xl,

. . .

rX,+k)K,(x,+,,...,xmi-1)

+(-l)kK,~,(~,,...,~,~l)K,~k~l(X,+,,..

. ,

%n+k+Z 1.

A ANSWERS TO EXERCISES 529

For example, we obtain the somewhat surprising noncommutative factoriza-

tions

(abc+a+c)(l

+ba)

(ab+l)(cba+a+c)

from the case k = 2, m = 0, n = 3.)

6.30 The derivative of K(xl , . . .

,x,)

with respect to xm is

K(x,,...

,xm-l)K(xm+l,...,xn),

and the second derivative is zero; hence the answer is

K(x,, . . .

,x,1

+K(xl,...,xm~1)K(x,+l,...,x,)~

6.31 Since xK =

(L)(n-

l)“k.

+ n

1)s =

(L)x”(n

I)*,

we have

]z]

ese

coefficients, incidentally, satisfy the recurrence

(n-l+k)/nkl~+~~~:/,

integersn,k>O.

6.32

Ek,,k{nlk}

= {“+,“+‘} and

&k$,,

{i}(m+l)”

= {G’,:}, both

of which appear in Table 251.

6.33 If n > 0, we have

[‘;I

i(n-

l)!

(Ht~

Hf-I,), by (6.71); {;} =

i(3”

3.2” + 3), by (6.19).

6.34 We have

(i’)

l/(k+

l),

(-,‘) =

Hr!,,

and in general

(z)

is given

by (6.38) for all integers n.

6.35 Let n be the least integer > l/e such that

[HnJ

[H,

-,J.

6.36 Now dk+, =

(lOOf(l

+dl)+...+(l+dk))/(lOO+k),

and the solution

is dk+l =

Hk+100

Hlol + 1 for k 3 1. This exceeds 2 when k 3 176.

6.37 The sum (by parts) is

H,,

+ . . . +

$-)

H,,

H,. The

infinite sum is therefore lnm. (It follows that

ym(k)

k>,

k(k+

mlnm,

m-l

because v,(k) = (m- 1)

xi,,

(k mod

mj)/mj.)

6.38

(-l)‘((“,‘)r-’

(1:

])Hk) + C. (By parts, using (5.16).)

6.39 Write it as

x,sjsn

jj’

xjbksn

and sum first on k via (6.67), to get

(n+l)HE-(2n+l)H,+2n.

530 ANSWERS TO EXERCISIES

6.40 If 6n

1 is prime, the numerator of

’

(-‘)k-

’

tjy-=

-Hzn

k=l

is divisible by 6n

1, because the sum is

Similarly if 6n + 1 is prime, the numerator of

x”,E,

(-1

)km

‘/k =

Han

~ Hln

is a multiple of 6n + 1. For 1987 we sum up to k = 1324.

6.41

‘&+I

(Lin+‘k+kl’LJ)

(L’“~k~‘2J),

hence we have Sn+l +

xk(

11*-i

kL’/2+11)

Sn+2.

The answer is F,+z.

6.42 F,,.

6.43 Set

in Ena0

F,z"

z/(1

z2)

to get

The sum is a

repeating decimal with period length 44:

0.1123595505617977ti28089887640449438202247191011235955+

6.44 Replace (m, k) by

(--m,

-k) or (k, -m) or (--k, m), if necessary, so

that m 3 k 3 0. The result is clear if m = k. If m > k, we can replace (m, k)

by (m

k, m) and use induction.

6.45 X, = A(n)oc+B(n)fi-tC(n)-y+D(n)&, where B(n) = F,, A(n) =

A(n) + B(n)

D(n) = 1, and B(n)

C(n) + 3D(n) = n.

6.46

$/2

and

-l/2.

Let

= cos 72” and v = cos 36”; then u =

2v2

1 and

v =

1-2sin'

18” =

1-2~‘.

Ijence

u+v

Z(u+v)(v-u),

and

4v2-2v-1

= 0.

We can pursue this investigation to find the five complex fifth roots of unity:

Q-1

i&Gfl

-Q

i&q

’

6.47

2Q5

&)”

&)n,

and the even powers of

cancel

"Let p be

any old

out. Now let p be an odd prime. Then (2kF,) = 0 except when k = (p

1)/2,

prime.”

;;d

(f,,!,)

0 except when k = 0 or k = (p

1)/2; hence

5(p

‘)/’ and

(See

j140/,

419.)

p+l

+5(Pm')/'

(mod p). It

can be shown that

5(PP’)/2

E 1 when

has

the form ‘Ok

and

5(P

’

:/2

E -1 when p has the form ‘Ok

6.48 This must be true because (6.138) is a polynomial identity and we can

set a, = 0.

A ANSWERS TO EXERCISES 531

6.49 Set

in (6.146); the partial quotients are 0,

2F3,

2F1,

2Fl, . .

(Knuth

[172]

noted that this number is transcendental.)

6.50 (a) f(n) is even

3\n. (b) If the binary representation of n is

(la'o"z...

lam

‘o”m )2,

where m is even, we have f(n) =

K(a’,az,.

,a, ‘).

6.51 (a) Combinatorial proof: The arrangements of

,2,.

. , p} into k sub-

sets or cycles are divided into “orbits” of 1 or p arrangements each, if we

add 1 to each element modulo p. For example,

{1,2,4I'J{3,51

{2,3,5IuI4,11

13,4,lIu{5,21

--f

14,5,2P~u,31

15,1,31w2,41

11,2,41~{3,51.

We get an orbit of size 1 only when this transformation takes an arrangement

into itself; but then k = 1 or k =

Alternatively, there’s an algebraic proof:

WehavexP-xfl-+xLandxE=xP-

x (mod p), since Fermat’s theorem tells

us that

x”

-x

is divisible by

(x-0)(x

1). . . (x

(p-l)).

(b) This result follows from (a) and Wilson’s theorem; or we can use

xp-’ E

xF/(x-1)

l(XP

-x)/(x- 1)

=xp

+xp-2 + .

..+x.

{“l’}

E [“:‘I

0 for 3 6 k 6

then {“12} E [pl’]

for 4 < k < p, etc. (Similarly, we have

[‘PpP’]

-{2ppm’}

E 1.)

(d) p!

=pP=

tk(-l)pPkpk[;]

=pP[~]-pP~'[pp,]+...+p3[~]~

P'[;]

+P[$

But

P[:]

P!,

[I]

P[I]

--P$]

+...+ppyP]

is a multiple of

p2.

(This is called Wolstenholme’s theorem.)

6.52 (a) Observe that

= Hk + HlnlP~/p, where Hz =

xc=,(k

p)/k.

(b) Working mod 5 we have H, = (0,

1,4,1,0)

for 0 <

< 4. Thus the first

solution is n = 4. By part (a) we know that

5\a,

5\ajn,sl;

so the next

possible range is n

20 + r, 0 <

6 4, when we have

&H4

H;e+~Hq+H,+~~=,

20/k(20+k).

The numerator of

H;,,

like the numerator

of HJ, is divisible by 25. Hence the only solutions in this range are n = 20

and n = 24. The next possible range is n = 100

now

&Hzo,

which is ;Hzo + H, plus a fraction whose numerator is a multiple of 5. If

$Hzo

m (mod 5), where m is an integer, the harmonic number

H’~o+~

will

have a numerator divisible by 5 if and only if m + H,

0 (mod 5); hence

m must be E 0, 1, or 4. Working modulo 5 we find $Hls =

&Hz0

+ &H4 3

-&Hh

= & E 3; hence there are no solutions for 100 6 n < 104. Similarly

there are none for 120 < n 6 124; we have found all three solutions.

(By exercise 6.51(d), we always have

p2\apP,,

p\apzPp,

and

p\a,r-,,

if p is any prime 3 5. The argument just given shows that these are the only

532 ANSWERS TO EXERCISES

solutions to

p\a,

if and only if there are no solutions to p

‘H,-l

+ H, E 0

(Attention, com-

(mod p) for 0 <

< p. The latter condition holds not only for p = 5 but

Puter

Programmers:

also for p = 13, 17, 23, 41, and 67-perhaps for infinitely many primes. The

Here’s an interest-

numerator of H,, is divisible by 3 only when n = 2, 7, and 22; it is divisible

ing condition to

test, for as many

by 7 only when n = 6, 42, 48, 295, 299, 337, 341, 2096, 2390, 14675, 16731,

Primes

asYou

can.)

16735, and 102728.)

6.53 Summation by parts yields

$$$(s((“-ZIHm+j

-1)-l).

6.54 (a) If m 3 p we have S,(p) E

S,m~,,P,,(p)

(mod p), since

’

= 1

when1

<k<p.

Also.&-l(p)=p-1

~-1.

IfO<m<p-1,wecanwrite

(b) The condition in the

h.int

implies that the denominator of Iln is not

divisible by any prime p;

he:nce

II,, must be an integer. To prove the hint, we

may assume that n > 1. Then

B2n

[(p-l

)\Vni]

‘E2

k=O

is an integer, by (6.78), (6.84), and part (a). So we want to verify that none

of the fractions (‘“,”

)Bkp2"

k/(2n + 1) = (2~)Bkp2”~k/(2n

k + 1) has a

denominator divisible by p. The denominator of (‘p)Bkp isn’t divisible by p,

since

has no

in its denominator (by induction); and the denominator

p2"

‘/(2n

k + 1) isn’t divisible by p, since 2n

k + 1 <

p'"

when

k 6

2n-2;

QED. (The numbers

1~~

are tabulated in

[185].

Hermite calculated

them through

1~s

in 1875

[153].

It turns out that

= Is =

110

112

= 1; hence there is actually a “simple” pattern to the Bernoulli

numbers displayed in the text, including

-$$(!).

But the numbers

I?,,

don’t

seem to have any memorable features when n > 6. For example,

BJ~

-86579

if,

and 86579 is prime.)

1 and

1 always divide 2n. If n is prime, the only

divisors of 2n are

2, n, and 2n, so the denominator of

BJ,,

for prime n > 2

will be 6 unless

2n+l

is also prime. In the latter case we can try

4n+3,

8n+7,

. . . ,

until we eventually hit a nonprime (since n divides 2”

‘n

+ 2”

’

1).

(This proof does not need the more difficult, but true, theorem that there are

infinitely many primes of the form 6k + 1.) The denominator of

BJ,,

can be 6

also when n has nonprime values, such as 49.

(The numerators of

Bernoulli numbers

have important

connections to

the known results

about Fermat’s

Last Theorem; see

Ribenboim j249j.j

A ANSWERS TO EXERCISES 533

6.55 The stated sum is

*(Xzn)(mn’;l)~

by Vandermonde’s convolution.

To get

(6.70)~

differentiate and set x = 0.

6.56 First replace k”+’ by ((k

m) + m) n + 1 and expand in powers of

m; simplifications occur as in the derivation of (6.72). If m > n or

m < 0, the answer is (-1 )“n!

m”/(“*“‘). Otherwise we need to take the

limit of (5.41) minus the term for k = m, as x

-m;

the answer comes to

(-l)nn!+(-l)m+l(~)mn(n+l

+mH,,

m-mHm).

6.57 First prove by induction that the nth row contains at most three

distinct values A,,

B, 3

C,;

if n is even they occur in the cyclic or-

der

[C,,,B,,A,,B,,C,],

while if n is odd they occur in the cyclic order

[C,,B,,,A,,A,,B,I.

Also

A2,+1

A2n

Bzn;

ALn

2A2n

B2n+1

B2n

+Cz,;

Bin =

A2n

+Bzn

CZn+l

2c>,;

1+C2n-1.

It follows that Q,,

F,+,.

(See exercise 5.75 for wraparound

binomial coefficients of order 3.)

6.58 (a)

n>OF;z'L

~(1

-z)/(l

+z)(l

-3z+t’)

;((2-3z)/(l

-32+

z2)

-2/(1

+zij.

(b)

naOF;t~n=z(l-2z-z2)/(1-4z-z2)(1+z-z2)=

;(2z/(l

-42-z2)+32/(1

+z-2’)).

(These formulas are obtained by squaring

or cubing Binet’s formula (6.123) and summing on n, then combining terms

so that

and $ disappear.) It follows that

Fi,,

4Ff,

3(-l)nF,.

(The corresponding recurrence for mth powers has been found by Jarden and

Motakin [163].)

6.59 Let m be fixed. We can prove by induction on n that it is, in fact,

possible to find such an x with the additional condition x $ 2 (mod 4). If x

is such a solution, we can move up to a solution modulo 3”+’ because

F8.3r,m,

3".

F8.3rIm~

3n+1

(mod

3”+‘);

either x or x +

8.3”

’

+ 16.3”

’

will do the job

6.60

+ 1,

and

1 are the only cases. Otherwise

the Lucas numbers of exercise 28 arise in the factorizations

F2m+(-lIm

= L,+,F,

F2m+1+(-lIm

= LnF,,,;

F2m-(-lIm

= L,

IF,+,;

F2m+1-(-lim

L+,F,.

(We have F,,,

(-l)nF,p,

L,F,

in general.)

6.61

1/F2,

= F, ,

/F,

FL,,-, /Fz,,, when m is even and positive. The

second sum is 5/4

F3.pp~/F3.21t,

for n 3 1.

534 ANSWERS TO EXERCISES

6.62 (a) A,, =

&?A,-,

A,-2

and B, =

&B,

B,~~2.

Incidentally,

we also have

&A,

= 2A,+, and

fig,,

A,, =

2B,

(b) A table of

small values reveals that

\/5F,,

even;

n odd.

(cl

WA,+1

B,-l/A,

= l/(Frn+l + 1) because

B,A,

B, rA,+l =

and

A,A,+l

&(Fz,+l

+ 1). Notice that

B,/A,+l

(F,/F,+r)[n

even] +

(L,/L,+li[n

odd].

(d)

Similarly, xi=,

1/(F2kmkl

= (Ao/BI

Al/B2) +

.. + (Anm~l/B,

A,/B,+j

) = 2

A,/B,+,

This quantity can also be

expressed as

(5F,/L,+l)

[n even] +

(L,/F,+,

) [n odd].

6.63 (a) [z]. There are [;‘-:I with n, = n and (n

l)[nk’] with n, < n.

(b)

(i).

Each permutation

. .

-1

of 11,. . , n-

leads to n permutations

n17t2..

.n,

1 n pj+l . . . on-l pi. If

1 has k excedances,

there are k+ 1 values of j that yield k excedances in

7~17~2

. . .

n,;

the remaining

n- 1~ k values yield k+ 1.

I-lence

the total number of ways to get k excedances

innln2...nnis(k+l)(“,‘)+((n-l)-(k-l))(z:;)=(t).

6.64 The denominator of (l’,‘) is 24nPvJini, by the proof in exercise 5.72.

The denominator of [

,I:‘,]

is the same, by (6.44), because

((i))

= 1 and

KS)

is even for k > 0.

6.65 This is equivalent to saying that (L)/n! is the probability that we

have

1x1

+ +

xnJ

= k, when

xl,

are independent random numbers

uniformly distributed between 0 and 1. Let

(x1

+ . . . +

) mod

Then

Yl,

f..,

y,, are independently and uniformly distributed, and

1x1

. . +

x,J

is the number of descents in the y’s. The permutation of the y’s is random,

and the probability of k descents is the same as the probability of k ascents.

6.66 We have the general formula

((;))

(2nr’){n~~:!iik}(-ll”.

for n > m > 0,

analogous to (6.38). When m = 2 this equals

((I)) =

{n:3}~-(2n+l){n~2}+

(‘“:‘){“:‘}

;3n+2

(2n + 3)2n+’

+i(4nz+6n+3).

6.67

~n(n+~)(n+l)(2H2n-H,)-&n(10n2+9n-1).

(It wouldbenice

to automate the derivation of formulas such as this.)

A ANSWERS TO EXERCISES 535

6.68 1 /k

/(k

+ z) = z/k2

z2/k3 + , and everything converges when

121

< 1.

6.69 Note that

nL=,

(1 + z/k)ePLlk =

(nnfL)nPLe(lnn

Hlllr.

If f(z) = &(z!)

we find

f(z)/z!

+ y q = H,.

6.70 For tan z, we can use tan

= cot

2 cot 22 (which is equivalent to the

identity of exercise 23). Also z/sin

= zcot

+ ztan

has the power series

&o(-l)“P’(4n

2)Bznz2”/(2n)!; and

tan2

In-

=In

-“”

lncosz

4n(4n-l)B2,~2n

(2n)(2n)!

4*(4" -2)BLnz2"

IL-‘)”

(2n)(2n)!

’

because

In sin

= cot

and

In cos

= -tan z.

6.71 Since tan2z

sec22 = (sin2 + cosz)/(cosz

sinz), setting x = 1 in

(6.94)

gives

T, (1) =

2nT,,

when n is odd,

T,,

(1) = 2"E,

when n is even, where

/cos

2 =

tn30

Ezn,

-‘“/(2n)’ (The

are called Euler numbers, not to be

confused with the Eulerian numbers (L)

6.72

2n+1(2n+'

l)B

,+i/(n

l),

if n > 0. (See (7.56) and (6.92); the

desired numbers are essentially the coefficients of 1

tanhz.)

6.73 cot(z +

rr)

= cot

and cot(z +

irr)

= -tan z; hence the identity is

equivalent to

cot

2gl

cot

3&z

)

k=O

which follows by induction from the case n = 1. The stated limit follows since

zcot

1 as

-+ 0. It can be shown that term-by-term passage to the limit

is justified, hence (6.88) is valid. (Incidentally, the general formula

cot

;

rcot

k=O

is also true. It can be proved from (6.88), or from

en2

k=O

which is equivalent to the partial fraction expansion of

l/(2”

l).)

536 ANSWERS TO EXERCISES

6.74 If p(x) is any polynomial of degree 6 n, we have

because this equation holds for x = 0, -1, . . . , -n. The stated identity is

the special case where p(x) =

xo,

(x) and x = 1. Incidentally, we obtain

a simpler expression for Bernoulli numbers in terms of Stirling numbers by

setting k = 1 in (6.99):

j-l)‘&

6.75 Sam Loyd [204, pages 288 and

3781

gave

the construction

and claimed to have invented (but not published) the 64 = 65 arrangement

in 1858. (Similar paradoxes go back at least to the eighteenth century, but

Loyd found better ways to present them.)

6.76 We expect

A,/A,-1

c$,

so we try

A,-,

= 618034 +

and A,,-2 =

381966-r. Then A,

=236068+2r, etc., and we find Am-is = 144-2584r,

A,,-19 = 154 +4181r. Hence

= 0, x = 154, y = 144, m = 20.

6.77 If

P(F,+j,

F,) = 0 for infinitely many

even.

values of n, then P(x,y) is

divisible by U(x,y)

1, where U(x,y) = x2

y2. For if t is the total

degree of P, we can write

P(XjYI

~4kXkY’pk

rj,kXjyk

Q(x,Y)

+R(x,y)

k=O

j+k<t

Then

P(Fn+l,Fn)

A ANSWERS TO EXERCISES 537

and we have xk=,

qk+k

= 0 by taking the limit as n

03.

Hence Q(x,y) is

a multiple of U(x,y), say A(x,y)U(x,y). But U(F,+,,F,) = (-1)” and n is

even, so

Pc(x,y)

= P(x,y)

(U(x,y)

l)A(x,y) is another polynomial such

that

Po(F,+,

, F,) = 0. The total degree of

is less than t, so

is a multiple

of U

1 by induction on t.

Similarly, P(x,y) is divisible by U(x,y) + 1 if

P(F,+,

,F,)

= 0 for in-

finitely many odd values of n. A combination of these two facts gives the

desired necessary and sufficient condition:

P(x,

y) is divisible by

U(x,

y)’

6.78 First add the digits without carrying, getting digits 0, 1, and 2. Then

use the two carry rules

O(d+l)(e+l)

lde,

O(df2)Oe

--t

ldO(e+l),

always applying the leftmost applicable carry. This process terminates be-

cause the binary value obtained by reading (b, . . .

bZ)F

as (b, . . .

b2)2

in-

creases whenever a carry is performed. But a carry might propagate to the

right of the “Fibonacci point”; for example,

)~+(l

becomes

(10.01)~.

Such

rightward propagation extends at most two positions; and those two digit po-

sitions can be zeroed again by using the text’s “add

algorithm if necessary.

Incidentally, there’s a corresponding “multiplication” operation on

nonnegative integers: If m =

Fj,

+. . . +

Fjq

and n =

Fk,

+. 3 . +

Fk,

in the

Fibo-

nacci number system, let m

n = xz=, EL=,

Fjb+k,,

by analogy with

mul-

Exercise: m

n =

tiplication of binary numbers. (This definition implies that m o n

x/!?

mn+ when m and n are large, although 1

+*n.) Fibonacci addition leads to

l(m+l)/@Jn+ a

proof of the associative law

o (m o n) =

o m) o n,)

ml(n+l

)/@I

6.79

Yes; for example, we can take

= 331635635998274737472200656430763;

A, = 1510028911088401971189590305498785 .

The resulting sequence has the property that A,, is divisible by (but unequal

to)

when n mod

rk,

where the numbers

(pk,

mk,rk) have the follow-

ing 18 respective values:

(3,4,1) (2,3,2) (5,5>1)

(7,833) (17,9,4)

(11,10,2)

(47,16,71

(19,18,10)

(61,15,3)

(2207,32,15)

(53,27,16) (31,30,24)

(1087,64,31)

(109,27,7) (41,20,10)

(4481,64,63)

(5779,54,52) (2521,60,60)

538 ANSWERS TO EXERCISIES

One of these triples applies to every integer n; for example, the six triples in

the first column cover every odd value of n, and the middle column covers all

even n that are not divisible by 6. The remainder of the proof is based on

the fact that A,,,+,, =

A,F,

1 +

A,+,F,,

together with the congruences

for each of the triples (pk, mk, rk). (An improved solution, in which

and

are numbers of “only” 17 digits each, is also possible [184].)

6.80 The matrix product is

(

zIxz,...,x,

Kn-1(~2,...,~,

1,x,)

)

I(xI,xL,...,x,-1.1

K~(x~,xz,...,x,~~,x,I

This relates to products of L and R as in (6.137), because we have

The determinant is K, (x1, , x,); the more general tridiagonal determinant

det

. . .

0 Y3 x3

1 :

. .

0 0

. . .

satisfies tl

recurrence

x,D,

ynD,-2.

6.81 Let

0~~~’

+ 1

/(al

l/(

))

be the continued fraction repre-

sentation of

‘. Then we have

aO+

l-z

Ao(z)

.lnaJ

)

TL31

(~1

A2(z)

where

Am(z) =

Z~4m.

z--q”,-l

z-qm--]

’

L(al,...,a,).

A proof analogous to the text’s proof of (6.146) uses a generalization of

Zeck-

endorf’s theorem (F’raenkel

[104,

$41).

If z = l/b, where b is an integer 3 2,

A ANSWERS TO EXERCISES

539

this gives the continued fraction representation of the transcendental number

‘1

tn3,

Lnaj,

as in exercise 49.

6.82 The sequences of exercise 62 satisfy A-,,, = A,,,,

B-~,,,

-B,,

and

A,&

Am+,

+A,-,;

A,&

Bm+n

‘L-n

;

BmB,

Am+,

-Am-n.

Let fk = Bmk/Amk+l and

= Amk/Bmk+,, where

i(n

m). Then

fkt-1

fk =

A~bn/(A2,k+,

AmI

and

gk+l

A~B,/(Azmk+n

A,);

hence we have

s;,,

FthLm

6.83 Let p = K(0,

al,

a2,. . . , a,), so that p/n is the mth convergent to the

continued fraction. Then

= p/n + (-1 )“‘/nq, where q = K(a, , . , a,,,, 6)

and

> 1. The points

{km}

for 0 6 k

n can therefore be written

0 1

n’

(-1)"~1

n-l +

(-l)mn,P1

. . . .

’

where

. .

X,-I

is a permutation of {1 , . . , n

l}. Let f(v) be the number

of such points < v;

t,hen

f(v) and vn both increase by 1 when v increases from

k/n to (k + 1 )/n, except when k = 0 or k = n

so they never differ by 2

or more.

6.84 By (6.139) and (6.136), we want to maximize

K(al,.

. . , a,,,) over all

sequences of positive integers whose sum is < n + 1. The maximum occurs

when all the a’s are 1, for if j 3 1 and a 3 1 we have

Kj+k+l(‘,...,

l,a+l,bl,...,

bk)

= Ki+lC+l(l

,...,

l,a,bl,...,

bk)+Kj(l,...,l)Kk(bl,...,bk)

Kj+k+l(l,...,l,a,bl,...,

bk)+Kj+k(l,...,

l,a,bl,...,

bk)

Kj+k+I(l,...,

l,o,bl,...,

bk).

(Motzkin and Straus

[220]

solve more general maximization problems on

con-

tinuants.)

540 ANSWERS TO EXERCISES

6.85 The property holds if and only if N has one of the seven forms 5k,

2.5k, 4.5k, 3j.5k, 6.5k, 7.5k, 14.5k.

6.86 A candidate for the case n mod 1 =

appears in

[179,

section

61,

although it may be best to multiply the integers discussed there by some

constant involving

fi.

6.87 (a) If there are only finitely many solutions, it is natural to conjec-

ture that the same holds for all primes. (b) The behavior of b, is quite

strange: We have b, = lcm( 1,. . . , n) for 968 6 n 6 1066; on the other hand,

Another reason to

b600

=kIIl(l,...

, 600)/(33 .52 .43). Andrew Odlyzko observes that p divides

remember

1066?

lcm( 1,. . .

,n)/b,

if and only if

kpm

6 n < (k +

1)~“’

for some m 3 1 and

some k < p such that p divides the numerator of Hk. Therefore infinitely

many such n exist if it can be shown, for example, that almost all primes

have only one such value of k (namely k = p

1).

6.88 (Brent

[33]

found the surprisingly large partial quotient 1568705 in

ey,

but this seems to be just a coincidence. For example, Gosper has found even

larger partial quotients in

rr:

The 453,294th is 12996958 and the 11,504,93lst

is 878783625.)

6.89 Consider the generating function tm,nZO ]“‘~“(w”‘z~, which has the

form ,Yn(wF(a,b,c)

+zF(a’,b’,~‘))~,

where F( a, b, c) is the differential op-

erator a +

b4,

+ ~4,.

7.1

Substitute

for

and

for o in the generating function, getting

1 /( 1

2’). This is like the generating function for T, but with

replaced

by 2’. Therefore the answer is zero if m is odd, otherwise Fm,2+l.

7.2 G(z) =

l/(1

22) +

l/(1

32); G(z) = ezr +

e3=.

7.3 Set

l/10

in the generating function, getting

7.4 Divide P(z) by Q(z), getting a quotient T(z) and a remainder

PO(Z)

whose degree is less than the degree of Q. The coefficients of T(z) must be

added to the coefficients

[z”]

Po(z)/Q(z) for small n. (This is the polynomial

T(z) in (7.28).)

7.5 This is the convolution of ( 1 + z’)~ with ( 1 +

z)~,

S(z) = (1

+z+z’+z3)‘.

Incidentally, no simple form is known for the coefficients of this generating

function; hence the stated sum probably has no simple closed form. (We can

use generating functions to obtain negative results as well as positive ones.)

A ANSWERS TO EXERCISES 541

I bet that the con-

troversial “fan of

7.6 Let the solution to

LX,

fi,

g,,

29,

(-1)“~

A(n)& +

B(n)13

+ C(n)y. The function 2” works when

= 1,

= 2,

y = 0; the function

(-1)”

works when

-1,

y = 0; the function

(-1)“n

works when

= 0,

= -1, y = 3. Hence A(n) + 2B(n)

= 2”,

A(n) -B(n) =

(-l)“,

and -B(n) + 3Cln) =

(-l)%.

7.7 G(z) = (z/(1

--z)‘)G(z)

+ 1, hence

G(z) =

1 ~ 22

-- +

=l+

;

-32+22

32 +

order zero” does

have one spanning

we have

Fzn

+ ‘n=Oj.

tree.

7.8 Differentiate (1

-x-l

twice with respect to x, obtaining

((H,,,

H,)’

(H$

HL2’))

Now set x = m.

7.9 (n +

l)(Hi

Hi2))

2n(H,

~ 1).

7.10 The identity

Hkm,,2-HP,,Z

+...+

2Hlk-Hk

implies that

(‘;)

(‘;

(2H2k

Hk) = 4nH,.

7.11 (a) C(z) =

A(z)B(z’)/(l

z). (b) zB’(z) = A(Zz)e’, hence A(z) =

‘l’B’($).

B(z)/(l

-z)‘+‘,

hence B(z) = (1

-z)‘+‘A(z)

and we

have fk(r) =

(‘l’)(-l)k.

7.12 C,. The nunibers in the upper row correspond to the positions of +1’s

in a sequence of

‘s

and -1

‘s

that defines a “mountain range”; the numbers

in the lower row correspond to the positions of -1’s. For example, the given

array corresponds to

7.13 Extend the sequence periodically (let

x,+k

Xk)

and define s, =

f...

+x,.

We have

~2~

= 21, etc. There must be a largest index

ki such that

Sk,

= j,

Sk,+,,,

j, etc. These indices

kl,

. . . , kl (modulo m)

specify the cyclic shifts in question.

For example, in the sequence

(-2,1,

-1

,O,

-1,

1,l)

with m = 10

and1=2wehavek,

=17,k2=24.

7.14 6 (z) =

-2zG(z)

e(z)2

(be careful about the final term!) leads

via the quadratic formula to

l+Zz-VTT-Q

G(z)

542 ANSWERS TO EXERCISES

Hence

gzn+l

= 0 and

gzn

(-1)“(2n)!C,.1,

for all

n>O.

7.15 There are

(L)b+k

partitions with k other objects in the subset con-

taining n + 1. Hence

B’

(z) =

eZB

(z). The solution to this differential equation

e(z)

eez+c,

and c = -1 since B(0) = 1. (We can also get this result by

summing (7.49) on m, since b, =

{t}.)

7.16 One way is to take the logarithm of

B(z) =

l/((l

-z)"'(l

-z2)"'(1

-~~)“~(l

-z4)aa

..).

then use the formula for In

and interchange the order of summation.

7.17 This follows since s,” tnePt dt = n

There’s also a formula that goes

in the other direction:

G(z) =

G ( zem~ie ) ee” d6 .

-x

7.18 (a)

<(z-

i);

(b) -L’(z); (c) L(z)/L(22). Every positive integer is

uniquely representable as m’q, where q is squarefree.

7.19

If n > 0, the coefficient

[zn]

exp(xln F(z)) is a polynomial of degree n

in x that’s a multiple of x. The first convolution formula comes from equating

coefficients of

Z”

F(z)"F(z)Y

F(z)~+Y.

The second comes from equating

coefficients of znP’ in

F'(z)F(z)"~'F(z)Y

F'(z)F(z)~+Y~',

because we have

F'(z)F(z)'-'

'i(F(z)")

x-'

nf,(x)znP’

ll>C

(Further convolutions follow by taking

a/ax,

as in (7.43).)

7.20 Let G(z) =

Ena

gnzn. Then

.zlGik)(z) =

nkg,,znPk+’ =

x(n

+ k

l)%Jn+kPlzn

lL)O lL>O

for all k,

3 0, if we regard

g,,

= 0 for n < 0. Hence if

PO(Z),

. . , P,(z) are

polynomials, not all zero, having maximum degree d, then there are polyno-

mials PO(n), . . . , p,,,+d(n) such that

mfd

Po(Z)G(Z)

+-+

P,(z)Giml(z) =

Pibhn+i-dZn.

n>o

j:=o

Therefore a differentiably finite G(z) implies that

m+d

pj(n+d)gn+j

for all n > 0.

j=O

This slow method of

finding the answer

is just the cashier’s

way of stalling until

the police come.

The USA has

two-cent pieces, but

they haven’t been

minted since 1873.

A ANSWERS TO EXERCISES 543

The converse is similar. (One consequence is that G(z) is differentiably finite

if and only if the corresponding egf,

e(z),

is differentiably finite.)

7.21 This is the problem of giving change with denominations 10 and 20, so

G(z)

l/(1

-z'")(l

-2") =

G(z"),

where

c(z)

l/(1

-z)(l

-2').

(a) The

partial fraction decomposition of

i;(z)

z)-~

-z)-'

(1 +

z)-'

[z”]

c(z)

i(2n

+ 3 + (-1)"). Setting n = 50 yields 26 ways to make

thepayment. (b) ~(z)=(1+z)/(l-z')2=(1+z)(l+2z2+3z4+~~~),~~

[z”]

c(z)

= Ln/2] + 1. (Compare this with the value

= Ln/5] + 1 in the

text’s coin-changing problem. The bank robber’s problem is equivalent to the

problem of making change with pennies and tuppences.)

7.22 Each polygon has a “base” (the line segment at the bottom). If A

and B are triangulated polygons, let AAB be the result of pasting the base

of A to the upper left diagonal of

and pasting the base of B to the upper

right diagonal. Thus, for example,

q LL=Q$

(The polygons might need to be warped a bit and/or banged into shape.)

Every triangulation arises in this way, because the base line is part of a unique

triangle and there are triangulated polygons A and B at its left and right.

Replacing each triangle by

gives a power series in which the coefficient

Z”

is the number of triangulations with n triangles, namely the number of

ways to decompose an (n+2)-gon into triangles. Since P = 1

+zP',

this is the

generating function for Catalan numbers

+ Cl z +

CZZ’

+ . . .

;

the number

of ways to triangulate an n-gon is

C,-2

= (2,1-,4)/(n

1).

7.23 Let a,, be the stated number, and b, the number of ways with a 2x 1 x 1

notch missing at the top. By considering the possible patterns visible on the

top surface, we have

2a,-l

+ 4b,_l + anPI + In = 01;

b, =

a,-1

b,-l.

Hence the generating functions satisfy A = 2zA + 4zB +

A +

B =

zB,

and we have

l-z

A(z)

(l+z)(l-4z+z2)'

This formula relates to the problem of 3 x n domino tilings; we have a,, =

U2,,

+Vzn+l +

(-l)n)

;(2+

fi)n+’

;(2-

fi)“+’

3(-l)“,

which is

(2 +

&)“+‘/6

rounded to the nearest integer.

544 ANSWERS TO EXERCISES

7.24

ntk,+...+k,rn

kl . . .

k,/m

= Fz,+l +

FznPl

2. (Consider the

coefficient

[znmll

&ln(l/(l

-G(z))), where G(z)

=z/(l

-z)~.)

7.25 The generating function is

P(z)/(l

z'~),

where

P(z)

2z2

. . + (m

)z+’

= ((m

)z”‘+’

mz”’ + z)/( 1

2)‘. The denominator

is Q(z) = 1

zm = (1

cu’z)(l

w'z)...(l

cumm’z). By the rational

expansion theorem for distinct roots, we obtain

nmodm =

k=l

7.26 (1

z2)5(z) =

F(z)

leads to

(2(n

l)F,

+ nF,+1)/5

as in

equation (7.60).

7.27 Each oriented cycle pattern begins with

or a 2 x k cycle (for

some k 3 2) oriented in one of two ways. Hence

Qn = Qn-I + Qn-2

+2Qnm

+2Qnp3

+...+ZQo

for n 3 2; Qo = QI = 1. The generating function is therefore

Q(z) =

zQ(z)

+z'Q(z)

+2z'Q(z)/(l

-z)+l

= l/(1-z-z2-222/(1-z))

-z)

(l-2z-2z2+23)

a2/5

=jqq+

c2/5

l-+2z

2/5

l+z'

and Q,, = (+2n+2 +

+~~2nP2

2(-1)“)/5

= (($‘l+l

$n+1)/&)2

Fi,,.

7.28 In general if A(z) = (1 +

...

zmp'

)B(z),

we have A, + A,,, +

rf2m + “’ = B(1) for 0 <

< m. In this case m = 10 and B(z) =

(1+z+~~~+z9)(1+z2+z4+z6+z8)(1+z5).

7.29 F(z)+ F(z)'+

F(z)~

. .

(l/(1

vmlJ8,

=z/(l-z-z2-z)=(l/(l-(l+&)z)-

so the answer is

((1 +

fi)"

fi)n)/J8.

7.30 XL=,

(2”nm’,Pk)

(anbnPk/(l

-olz)k+anPkbn/(l

-@z)~), by exercise 5.39.

7.31 The dgf is <(~)~/<(z-l); hence we find g(n) is the product of (k+l-kp)

over all prime powers

that exactly divide n.

7.32 We may assume that each

3 0. A set of arithmetic progressions

forms an exact cover if and only if

Zbl

Zb,

1-z

l-zo-1

+ . +

Za, .

A ANSWERS TO EXERCISES 545

Subtract zbm/( 1

z”m

) from both sides and set

= e2xi/am. The left side is

infinite, and the right side will be finite unless a,,-1 = a,.

7.33

(-l)n--m+‘[n>m]/(n

m).

7.34 We can also write G,(z) =

tk,+,m+,,k,+,=n

(kl~~k+~+l)(zm)km+l.

general, if

we have

G,_., + z2GnP2

. . +

zrGnmmr

+ [n=O], and the generating

function is

l/(

zzw2

. . .

z,.w’). In the stated special case the

answer is

l/(1

-w

zmwm+‘). (See (5.74) for the case m = 1.)

7’35

Cal

t0<k<n

(l/k+l/(n-k)) =

iH,-l.

(b)

[z”]

(ln&)‘=

$[i]

$H,-l

by (7.50) and (6.58). Another way to do part (b) is to use the rule

[zn]

F(z) = i[znP’] F’(z) with F(z) = (In

&)2.

7.36

+$A(P).

7.37 (a) The amazing identity azn =

az,+l

= b, holds in the table

nlO1234

5 6 7 8 910

1 1

4 4

6 6 10 10 14

4 6 10 14

20 26 36 46 60 --?J

(b) A(z) =

l/((l

z)(l -z')(l

z4)(1

-z8)...). (c) B(z) = A(z)/(l -z),

and we want to show that A(z) = (1 + z)B(z’). This follows from A(z) =

A(z’),‘(

z).

7.38 (1

wz)M(w,z)

tm,n3,(min(m,n)

min(m-l,n-l))wmzn

m,n

wmzn = wz/( 1

w) (1 ~ z). In general,

M(z,,...,z,)

..zm

(1 -z,)...(l

-z,)(l

-z1

. ..z.) .

7.39 The answers to the hint are

ak,

ak2

ak,,

and

ok,

ak2

ak,,,

l<k,<kl<...<k,,,$n

l<k,$k>s...$k,,,$n

respectively. Therefore: (a) We want the coefficient of zm in the product

(l+z)(l+2z)...(l+nz).

Thisisthereflectionof(z+l)“,soitis[~~~]+

[“,“]Z+

.‘.

+’

[“:‘]z”

and the answer is

[,T:

lm]. (b) The coefficient of

z”’

l/((l

-z)(l

-2z)...(l

-nz))

is {“‘,‘“} by(7.47).

546

ANSWERS TO EXERCISES

7.40 The egf for (nF, 1

F,) is (z

lip(z)

where

i(z)

J&0

F,z”/n! =

(e@’

e&‘)/fi.

The egf for (ni) is e

-‘/(l

z). The product is

l/2

112

-el@

‘12)

liqe

-42

6’).

We have

i(z

-?(-z).

So the answer is

(--l)"F,.

7.41 The number of up-down permutations with the largest element n in

position 2k is

(z”k

‘,)A

,A,,

-2k.

Similarly, the number of up-down permu-

tations with the smallest element 1 in position 2k + 1 is (‘&‘)A2kAn

because down-up permutations and up-down permutations are equally nu-

merous. Summing over all possibilities gives

AkA,~r~~k+2[n=O] +

[n=l]

The egf

therefore satisfies 2A’(2) = A(z)’ + 1 and A(0) = 1; the given

function solves this differential equation.

7.42 Let a, be the number of Martian DNA strings that don’t end with

or e; let b, be the number that do. Then

3~~1

2b,-

1 + [n = 01, b,

h-1

;

A(z) = 3zA(z)

+2zB(z)

+ 1 ,

B(z)

ZzA(z)+zB(z);

and the total number is [z”]

(

1 +

z)/(

z2)

F3,,

+2.

7.43 By (5.45),

g,,

= An6(0). The nth difference of a product can be

written

A”A(z)B(z) =

‘k”

(ACED

kA(Z))(AR

k~(~))

)

and

= (1 +

A)nPk

(“j “)Ai. Therefore we find

(E)

(“yk)

fjfkgn-k.

This is a sum over all trinomial coefficients; it can be put into the more

symmetric form

fj+kgk+L.

A ANSWERS TO EXERCISES 547

The empty set

is pointless.

7.44 Each partition into k nonempty subsets can be ordered in k! ways, so

= k!. Thus

Q(Z)

= ,&k>O {E}k!zn/n! = tkao(eZ

l)k =

l/(2

e’).

And this is the geometric series

tkao

ekZ/2k+‘,

hence

= 1

/2kf’.

Finally,

= 2k; consider all permutations when the x’s are distinct, change each ‘>’

between subscripts to ‘<’ and allow each ‘<’ between subscripts to become

either ‘<’ or ‘=‘. (For example, the permutation

~3x2 produces

< x3 < x2

and

= x3 < x2, because 1 < 3 > 2.)

7.45 This sum is

,&,

r(n)/n2,

where r(n) is the number of ways to write

n as a product of two relatively prime factors. If n is divisible by t distinct

primes, r(n) = 2t. Hence r(n)/n2 is multiplicative and the sum is

l+l+F

p4-)

I$+&)

n(g)

L(2)2/L(4)

7.46 Let

= ,Y0skSn,2 (ni2k)

CX~.

Then

1 + as,,

+ [n =O], and

the generating function is

l/(1

az3). When a =

-A,

the hint tells us

that this has a nice factorization

l/(1

iz)(

52)‘. The general expansion

theorem now yields

($n+c)($)n+$(-i)n,

and the remaining constant c

turns out to be $.

7.47 The Stern-Brocot representation of

R(LR2)D0,

because

d3+1=2+

'+x&+1

The fractions are

3, 5,

. .

;

they eventually have the cyclic

pattern

Vzn-1

+v2,+

U2n+VZn+l u2n+2+v2n

VZn+l

+VZn+3

U2n

VZn+l

UZn+VZn+l

U2n+2

. . . .

7.48 We have

= 0, and if

= m the generating function satisfies

aG(z)+bzP’G(z)+czP2(G(z)-mz)+&

= 0.

Hence G(z) =

P(z)/(az'

+ bz + c)(l

-z)

for some polynomial P(z). Let

and

be the roots of

cz2

+ bz + a, with

lp,

( >

/pal.

4ac 6 0 then

lp,

1’

= a/c is rational, contradicting the fact that

approaches

548 ANSWERS TO EXERCISES

Hence

= (-b +

dm)/Zc

= 1 +

and this implies that

-c,

b =

-2c,

= 1

fi.

The generating function now takes the form

z(m

(r + m)z)

G(z)

= (1 -22-z2)(1 -2)

-r

(m +

2r)z

= 2(1-22-z')

+2(1-z!

= mz+

(2m-r)z2

+...

where

= d/c. Since

is an integer,

is an integer. We also have

a(1

+Jz)n

+a(1

-Jz)n+

tr =

[cx(l+Jz)"],

and this can hold only if

= -1, because (1

a)”

alternates in sign as

it approaches zero. Hence (a, b, c, d) = *(

1,2,

-1,l).

Now we find

(1 +

m), which is between 0 and 1 only if 0 < m 6 2. Each of

these values actually gives a solution; the sequences (g,,) are

(O,O,

1,3,8,.

.),

(0,1,3,8,20

,...

), and

(0,2,5,13,32

,...

7.49 (a) The denominator of

(l/(1

fl)z)

l/(1

&!)z))

z2;

hence a,, =

2a,-l

a,-2

for n 3 2. (b) True because a,, is even

and -1

0. (c)

Let

(!?+!t?)n+(v)n

We would like b, to be odd for all n > 0, and -1 < (p

Jsi)/Z

< 0. Working

as in part (a), we find

= 2,

and b, = pb,~-l +

i(q

p2)bn.

for

n 3 2. One satisfactory solution has p = 3 and q = 17.

7.50 Extending the multiplication idea of exercise 22, we have

Q =

QAQ

QfjQ

QQdQ

+..

Replace each n-gon by

znm

‘.

This substitution behaves properly under mul-

tiplication, because the pasting operation takes an m-gon and an n-gon into

an (m + n

2)-gon. Thus the generating function is

Q =

1+zQ2+z2Q3+z3Q4+...

I.+*

-zQ

and the quadratic formula gives Q = (1

+z-dl

6z +

)

/2z.

The coefficient

of z” -’ in this power series is the number of ways to put nonoverlapping

diagonals into a convex n-gon. These coefficients apparently have no closed

Give

Legen-

form in terms of other quantities that we have discussed in this book, but

their asymptotic behavior is known

[173,

exercise 2.2.1-121.

~~~~$~j~~~~

closed form.

A ANSWERS TO EXERCISES

549

Incidentally, if each n-gon in Q is replaced by wzne2 we get

l+z-dl-(4w+2)z+z2

211

+w)z

a formula in which the coefficient of wmzn

’

is the number of ways to divide

an n-gon into m polygons by nonintersecting diagonals.

7.51 The key first step is to observe that the square of the number of ways

is the number of cycle patterns of a certain kind, generalizing exercise 27.

These can be enumerated by evaluating the determinant of a matrix whose

eigenvalues are not difficult to determine. When m = 3 and n = 4, the fact

that cos36’ =

$/2

is helpful (exercise 6.46).

7.52 The first few cases are

PO(Y)

= 1,

pi(y)

= y,

= y2 + y,

P3(Y)

= Y3 +

3Y2

+ 3Y.

Let p,(y) = qrn(x) where y = x(1 ~ x); we

seek a generating function that defines qln+l (x) in a convenient way. One

such function is

x:,

q,(x)z”/n!

= 2eixZ/(eiZ +

l),

from which it follows that

q,,(x) =

i”E,

(x), where E,(x) is called an Euler polynomial. We have

t(-l)“x”bx

$(-1)

‘+’ E,(x), so Euler polynomials are analogous to Ber-

noulli polynomials, and they have factors analogous to those in (6.98). By

exercise 6.23 we have nEnpl (x) =

EL,

(z)Bkx”

k(2-2k+‘);

this polynomial

has integer coefficients by exercise 6.54. Hence

q2,,

(x), whose coefficients

have denominators that are powers of 2, must have integer coefficients. Hence

p,(y) has integer coefficients. Finally, the relation (4y

l)pl(y)

+ 2pk(y) =

Ln(2n

)p,-l

(y) shows that

2m(2m-1)

n =

m(m+l)

I I

, +

2n(2n

n-l

I I

m-l

’

and it follows that the

I:[’

s are positive. (A similar proof shows that the

related quantity

(-l)n(2n

+ 2)Eln+l (x)/(2x

1) has positive integer coeffi-

cients, when expressed as an nth degree polynomial in y.) It can be shown

that

I;‘1

is the Genocchi number

(-l)np’

(22n+’

~ 2)B2, (see exercise 6.24),

and that

Inn,1

(;),

Inn21

= 2(“qf’)

+3(T),

etc.

7s53

P(l+V~~,-,+V4,,r~1/6.

Thus, for example,

TJO

PJL

= 210;

T2s5

PI65

= 40755.

7.54 Let

be the operation on power series that sets all coefficients to zero

except those of zn where n mod m = k. The stated construction is equivalent

to the operation

EoSEoS(Eo+El)S

. . .

S(Eo+E,

+...+E,,m,)

550 ANSWERS TO EXERCISES

applied to

l/(

z), where S means “multiply by

l/(

z).” There are m!

terms

EosEk,

s&s

. . .

SEk,,,

where 0 6 ki < j, and every such term evaluates to zrm/( 1

zm)

m+’

is the

number of places where ki < ki+r .

Exactly (y) terms have a given value of r,

so the coefficient of zmn is

x2;’

(~)(“‘,‘~‘)

(n+l)“’

by (6.37). (The fact

that operation

can be expressed with complex roots of unity seems to be

of no help in this problem.)

7.55 Suppose that Po(z)F(z) +

...

P,(z)Fiml(z)

= Qo(z)G(z)

...

Qn(z)Gin)(z) = 0, where P,(z) and Qn(z) are nonzero. (a) Let H(z) =

F(z) + G (2). Then there are rational functions

Rk,l

(z) for 0 <

< m + n such

that Hck)(z) =

Rk,o(z)FCo)(z)

...

+ Rk,mpl(~)F’mp’)(~) + Rk,,,(z)Gcol(z) +

. .

Rk,m+n-1 (z) G

(n-‘i(z). The

m+n+l

vectors

(Rk,O(z),...,Rk,m+n-l(~))

are linearly dependent in the (m + n)-dimensional vector space whose com-

ponents are rational functions; hence there are rational functions

S(z),

not

all zero, such that

SO(Z)H~~~(Z)

...

+ S,+,(Z)H~~+~~(Z) = 0. (b) Sim-

ilarly, let H(z) = F(z) G (2). There are rational Rk,l(z) for 0 $

< mn

with H’k’(~) =

XL;’

I;<:

Rk,ni+j

(z)Fiii(z)Gii)(z),

henceSo(z)H”‘(z)+...+

Smn(z)Himnl (z) = 0 for some rational St(z), not all zero. (A similar proof

shows that if (fn) and

(gn)

are polynomially recursive, so are (f, +

gn)

and

(fngn). Incidentally, there is no similar result for quotients; for example, cos z

is differentiably finite, but 1 /cos z is not.)

7.56 Euler showed, incidentally, that this number is also

[zn]

l/d-,

and he gave the formula a, =

tk,O

n&/k.

1’

He also discovered a “memorable

failure of induction” while examining these numbers: Although

3a,

an+1

equal to

F,-

(F,-,

+ 1) for 0 $ n < 9, this empirical law mysteriously breaks

down when n is 9 or more!

7.57 (Paul ErdGs currently offers $500 for a solution.)

8.1

& + & +

+ & + & + & =

(In fact, we always get doubles with

probability

when at least one of the dice is fair.) Any two faces whose sum

is 7 have the same probability in distribution

Prl,

so S = 7 has the same

probability as doubles.

8.2

There are 12 ways to specify the top and bottom cards and

50!

ways to

arrange the others; so the probability is 12.50!/52! = 12/(51.52)

8.3

&(3+2-t..

.+9+2)

= 4.8;

$(32+22+.

.+92+22-10(4.8)2)

8.6.

The true mean and variance with a fair coin are 6 and 22, so Stanford had

an unusually heads-up class. The corresponding Princeton figures are 6.4 and

A ANSWERS TO EXERCISES 551

562

12.5. (This distribution has ~4 = 2974, which is rather large. Hence the

standard deviation of this variance estimate when n = 10 is also rather large,

J2974/10 +

2(22)2/9

20.1 according to exercise 54. One cannot complain

that the students cheated.)

a.4

This follows from (8.38) and (88g), because

F(z)

= G(z)H(z). (A

similar formula holds for all the cumulants, even though

F(z)

and

G(z)

may

have negative coefficients.)

8.5

Replace H by p and T by q = 1

-p.

we have

p2qN

and

pq2N

= iq +

the solution is p =

l/~$~,

q = l/G.

8.6

In this case

Xly

has the same distribution as X, for all y, hence

E(XIY) = EX is constant and

V(E(XlY))

= 0. Also

V(XlY)

is constant and

equal to its expected value.

8.7

We have 1 =

(PI

+pz+.+.+ps)’ 6 6(p~+p~+.~.+p~) by Chebyshev’s

summation inequality of Chapter 2.

8.8

Let p =

Pr(wEAflB),

q = Pr(~ueA), and r = Pr(w$B). Then

p+q+r=l,andtheidentitytobeprovedisp=(p+r)(p+q)-qr.

8.9 This is true (subject to the obvious proviso that

and G are defined

on the respective ranges of X and Y), because

Pr(F(X)=f and G(Y)=g) =

Pr(X=xandY=y)

Y&F-‘(~)

YEG-‘(9)

Pr(X=x) .Pr(Y=y)

c&F-‘(f)

YEG-‘(9)

Pr(F(X)

=f) . Pr(G(y) = g) .

8.10 Two. Let

be medians; then 1 < Pr(X<xl) + Pr(X>xz) <

hence equality holds. (Some discrete distributions have no median ele-

ments. For example, let

be the set of all fractions of the form

&l/n,

with

Pr(+l/n) = Pr(-l/n) =

$n2.)

8.11 For example, let K = k with probability 4/( k + 1) (k + 2) (k +

3))

for all

integers k 3 0. Then EK = 1, but

E(K’)

00.

(Similarly we can construct

random variables with finite cumulants through

but with

K,+I

= co.)

8.12

(a) Let

= Pr(X = k). If 0 < x < 1, we have Pr(X 6

= tk<,.pk <

k<r

xk-r,,k

xkxk-’

= x-‘P(x). The other inequality has a similar

proof. (b) Let x = a/(1

a) to minimize the right-hand side. (A more

precise estimate for the given sum is obtained in exercise 9.42.)

552 ANSWERS TO EXERCISES

8.13 (Solution by Boris Pittel.) Let us set Y =

(XI

+ . . . + X,)/n and

Z = (X,+1 + . + X2,)/n. Then

Pr(JZ-cxJ

IY-al)

The last inequality is, in fact, ‘>’ in any discrete probability distribution,

because Pr(Y = Z) > 0.

8.14 Mean(H) = pMean(F) + qMean(G); Var(H) = pVar(F) + qVar(G) +

pq(Mean(F)-Mean(G))‘. (A mixture is actually a special case of conditional

probabilities: Let Y be the coin, let

XIH

be generated by F(z), and let

XIT

be generated by G(z). Then VX = EV(XIY) + VE(XlY), where EV(XlY) =

pV(XIH) + qV(XIT) and VE(XlY) is the variance of

pzMeanCF)

qzMeanfG).)

8.15 By the chain rule, H’(z) =

G’(z)F’(G(z));

H”(z) =

G”(z)F’(G(z))

G’(z)‘F”(G(z)). Hence

Mean(H) = Mean(F) Mean(G) ;

Var(H) = Var(F) Mean(G)’ + Mean(F) Var(G)

(The random variable corresponding to probability distribution H can be un-

derstood as follows: Determine a nonnegative integer n by distribution F;

then add the values of n independent random variables that have distribu-

tion G. The identity for variance in this exercise is a special case of (8.105),

when X has distribution H and Y has distribution F.)

8.16

ewizmm’)/(l

-w).

8.17

Pr(Y,,,

6 m) =

Pr(Y,,,

+ n 6 m + n) = probability that we need <

+ n tosses to obtain n heads = probability that m + n tosses yield 3 n

heads = Pr(X

m+n,p

3 n). Thus

n+k-1

pnqk =

(m;n)p*qm+n~k

k>n

(m;n)pm+n-kqk;

k<m

and this is (5.19) with n =

I-,

= q, y = p.

8.18 (a) Gx(z) =

epfz

‘1.

(b) The mth cumulant is

for all m 3 1. (The

case p = 1 is called

in (8.55).)

A ANSWERS TO EXERCISES 553

8.19 (a)

Gx,+x,(z)

Gx,

(z)Gx>(z) = ervl+plrir ‘I. Hence the probability

eW1+p2

(~1

+ pz)“/n!; the sum of independent Poisson variables is Poisson.

(b) In general, if K,X denotes the mth cumulant of a random variable X, we

have

K,(aX,

bX2‘t

= am(K,X1) + bm(K,Xz), when a, b 3 0. Hence the

answer is 2mpl + Smu2.

8.20 The general pgf will be G(z) = zm/F(z), where

F(z) = zm

-r)~aikl;A:kl=A,kl]Zm~~k,

k=l

F’(1)

m-f&,~[A(k’=A~k,],

k=l

F”(1) =

m(m-

-2~(m-k)A:kI[Aik)=Alkl].

k=l

8.21 This is Ena0

qn,

where

q,,

is the probability that the game between

Alice and Bill is still incomplete after n flips. Let

be the probability that

the game ends at the nth flip; then p,, + q,, =

q,-l.

Hence the average time

to play the game is

Ena,

wn=(q0-q41)+4ql

-q2)+3(42-43)+...=

. = N, since

lim,,,

nq, = 0.

Another way to establish this answer is to replace H and T by

:z.

Then the derivative of the first equation in (8.78) tells us that N (1) + N’( 1) =

N’(l)+S;(l)+Sf,(l).

By the way, N =

8.22 By definition we have V(XjY) = E(X’lY)

(E(XlY))’ and V(E(xlY)) =

E(KW’l12)

(E(EIW’)))2;

ence

E(V(XlY))

+V(E(XlY))

= E(E(X’lY))

(E(E(X/Y)))‘. But E(E(XlY)) = EX and E(E(X’IY)) = E(X2), so the result is

just VX.

8.23

Let C& =

{

}’

and

fll

=(m,

H}‘;

and let

be the

other 16 elements of n. Then Prl1 (w)

Pro0

(cu)

according

as w

no,

RI,

02.

The events A must therefore be chosen with

elements

fromnj,

where

(ko,kl,kl)

is one ofthe following:

(O,O,O),

(0,2,7),

(0,4,14),

(1,4,4),

(1,6,11),

(2,6,1),

(2,&g),

(3,&15),

(3,10,5),

(3,12,12),

(4,12,2),

(4,14,9),

(4,16,16).

Forexample, thereare

($(‘,“)(‘f)

eventsoftype

(2,6,1).

The total number of such events is

[z’]

(1 + z”)~ (1 +

‘)16( 1 + z2)16, which

turns out to be 1304927002. If we restrict ourselves to events that depend

on S only, we get 40 solutions S

E’

A, where A =

{

f2,

, t},

{

f2, 5,9},

(2,12

PO,

z, 5,9}, {2,4,6,8,10,12},

{

f, ,7, z, 4,

lo},

and the complements of

these sets. (Here the notation

‘,:’

means either 2 or 12 but not both.)

554

ANSWERS TO EXERCISES

8.24 (a) Any one of the dice ends up in J’s possession with probability

;

I,“+‘;$+;

p, hence p = A. Let q = A. Then the pgf for J’s total holdings

with mean (2n +

l)p

and

(b)

(;)p3q2

(;)p;'4

(;)p'

variance (2n +

l)pq,

by (8.61).

.585.

8.25 The pgf for the current stake after n rolls is G,(z), where

Go(z)

z*;

G,(z)

I;=, G,

,(~"~~')9/6,

for n > 0.

This problem can

(The noninteger exponents cause no trouble.) It follows that Mean(G,,) =

perhaps

so’ved

Mean(G,

and

Var(G,)

-t

Mean(

= f$(Var(G,,

+ Mean(G,

1’).

more easily without

So the mean is always A, but the variance grows to

((g)”

~ l)A2.

geprat;ngfunct;ons

than with them.

8.26 The pgf

FL,,(Z)

satisfies F,',,,(z) =

FL,,

L(z)/L; hence Mean(Ft,,) =

FL,,

(1) = [n 3 L]/l and

F,lln

(1) = [n 3

21]/L2;

the variance is easily computed.

(In fact, we have

which approaches a Poisson distribution with mean

l/1

as n

co.)

8.27 (n2L3

3nZ21-1

Zt:)/n(n

l)(n

2) has the desired mean, where

+ +

Xk.

This follows from the identities

E(x211)

= np3

+n(n-

1)p2p1

;

E(x:)

w3+3n(n-1)p2pl

+n(n-l)(n-2)p:.

Incidentally, the third cumulant is

= E( (X-EX)3), but the fourth cumulant

does not have such a simple expression; we have

= E (( X

EX)4)

VX)'.

8.28 (The exercise implicitly calls for p = q

but the general answer is

given here for completeness.) Replace H by pz and T by qz, getting S*(z) =

p2qz3/(1

-pz)(l

qz)(l

-pqz’)

and

SB(Z)

pq2z3/(1

qz)(l

-pqz2).

The

pgf for the conditional probability that Alice wins at the nth flip, given that

she wins the game, is

S*(Z)

1-Pq

SAtI)

l-pz

l-qz

l-pqz2

This is a product of pseudo-pgf’s, whose mean is

3+p/q+q/p+2pq/(

1 -pq).

The formulas for Bill are the same but without the factor

q/(

-pz),

so Bill’s

mean is 3 + q/p +

2pq/(l

-pq). When p = q =

the answer in case (a) is

A ANSWERS TO EXERCISES 555

in case (b) it is

Bill wins only half as often, but when he does win he

tends to win sooner. The overall average number of flips is

+ $. $! = l$,

agreeing with exercise 21. The solitaire game for each pattern has a waiting

time of 8.

8 29

Set H = T =

l+N(H+T)

N+SA+SB+S~

N HHTH = SA(~+HTH)+S~(HTH+TH+~)+S~(HTH+TH)

N HTHH =

SA(THH

+ H) +

S~(THH

+ 1) +

S~(THH)

N THHH =

SA(~H)

to get the winning probabilities. In general we will have

+ Ss +

= 1

and

SA(A:A)

+ Ss(B:A) + Sc(C:A) = SA(A:B) +Ss(B:B) + Sc(C:B)

= SA(A:B) + Ss(B:C) + Sc(C:C).

In particular, the equations 9SA+SSs+3Sc =5SA+9Ss+Sc = 2SA+4Ss+9Sc

imply that

Sc =

8.30 The variance of P(hl , . . .

h,;

k) I k is the variance of the shifted bino-

mial distribution ((m ~ 1 +

z)/m)

’

z, which is

(k-1)($)(1

by (8.61).

Hence the average of the variance is Mean(S)(m

1)/m’.

The variance of

the average is the variance of (k

)/m,

namely Var(

S)/m’.

According to

(8.105), the sum of these two quantities should be VP, and it is. Indeed, we

have just replayed the derivation of

(8.~5)

in slight disguise. (See exercise 15.)

8.31 (a) A brute force solution would set up five equations in five unknowns:

A = ;zB +

;zE;

B =

;zC;

C = 1 + +zB +

;zD;

D =

;zC

;zE;

E =

;zD.

But positions C and D are equidistant from the goal, as are B and E, so we

can lump them together. If X = B + E and Y = C + D, there are now three

equations:

A =

;zX;

X =

;zY;

Y = 1

+;zX+;zY.

Hence A = z2/(4 ~ 22

z’); we have Mean(A) = 6 and Var(A) = 22. (Rings

a bell? In fact, this problem is equivalent to flipping a fair coin until get-

ting heads twice in a row: Heads means “advance toward the apple” and

tails means “go back.“) (b) Chebyshev’s inequality says that Pr(S 3 100) =

Pr( (S

6)’ 3 942) 6 22/942

.0025. (c) The second tail equality says that

Pr( S > 100) < 1 /x98 (4

x2) for all x 3

and we get the upper bound

0.00000005 when x =

(v’m

99)/l 00. (The actual probability is approx-

imately 0.0000000009, according to exercise 37.)

556 ANSWERS TO EXERCISES

8.32 By symmetry, we can reduce each month’s situation to one of four

possibilities:

“Toto, I have a

feeling we’re not in

D, the states are diagonally opposite;

A, the states are adjacent and not Kansas;

K, the states are Kansas and one other;

the states are the same.

Kansas anymore.

”

-Dorothy

Considering the Markovian transitions, we get four equations

D = 1

+z($D++)

A =

z(;A+

AK)

4 4 4

K =

z(~D+~A+~~K)

S =

z(fD

+ ;A +

$K)

whosesumisD+K+A+S=l+z(D+A+K).

Thesolutionis

812-452=

423

243-243z+24z2 +8z3

but the simplest way to find the mean and variance may be to write z = 1 + w

and expand in powers of w, ignoring multiples of w2:

g+‘593w+.,..

16 512

?+?115w+..,.’

8 256

fi+&i!iiw+.,..

8 256

Now

S’(1)

and

is”(l)

$$!

The

mean is

and the variance is

(Is there a simpler way?)

8.33 First answer: Clearly yes, because the hash values

hl,

. . . , h, are

independent. Second answer: Certainly no, even though the hash values

hi,

h, areindependent. WehavePr(Xj=O)

=xt=,

sk([j#k](m-1)/m)

(;‘lsj)(m-1)/m,

but

Pr(XI=Xz=O)

=xL=,

sk[k>2](m-l)2/m2

= (1

sl)(m-

1)=/m= # Pr(XI =0) Pr(X2 =O).

8.34 Let [z”] S,(z) be the probability that Gina has advanced < m steps

after taking n turns. Then

S,(

1) is her average score on a par-m hole;

[z”‘]

S,(z) is the probability that she loses such a hole against a steady player;

and 1

[z’“~~‘]

S,(z) is the probability that she wins it. We have the recurrence

So(z) = 0;

S,(z)

+ P&l~~Z(~) +

q&lpl

(z))/(l

=I,

for m > 0.

A ANSWERS TO EXERCISES 557

To solve part (a), it suffices to compute the coefficients for

m,n

< 4; it is

convenient to replace

by 100~ so that the computations involve nothing

but integers. We obtain the following tableau of coefficients:

so0

0 0 0 0

64 256

95 744 4432

23552

100

9065 104044 819808

.Sq

100 9975 868535 12964304

Therefore Gina wins with probability 1

.868535 =

.131465;

she loses with

probability .12964304. (b) To find the mean number of strokes, we compute

S,(l) =

Sz(1)

= gj$; SJ(1) =

S‘$(l)

J#gg.

(Incidentally,

(

E 4.9995; she wins with respect to both holes and strokes

on a par-5 hole, but loses either way when par is 3.)

8.35 The condition will be true for all n if and only if it is true for n =

by the Chinese remainder theorem. One necessary and sufficient condition is

the polynomial identity

but that just more-or-less restates the problem. A simpler characterization is

(P2+P4+P6)(P3+P6)

p6,

(PI

+P3+P5)(PZ+pS)

?-75,

which checks only two of the coefficients in the former product. The general

solution has three degrees of freedom: Let

and

put

=alh,

=aobz,

= alhI

= aoh,

= alh,

= aobo.

8.36 (a)

q q q q q q

. (b) If the kth die has faces with ~1,

. . . 9

spots, let pk(Z) = 2”

. .+I?“. We want to find such polynomials with

Pl(Z)...P,(Z)

(z+z2+z3+z4+z5+z

The irreducible factors of this

)

polynomial with rational coefficients are zn(z + 1 )“(z’ + z + 1 )“(z’

z +

1)";

hencep~(z)mustbeoftheformzak(z+l)b~(z2+z+1)Ck(z2-z+l)d~.

must have

since pk(0) = 0; and in fact

since

a,, = n.

Furthermore the condition pk( 1) = 6 implies that

It is now easy

to see that 0 6

< 2, since

> 2 gives negative coefficients. When d = 0

and d = 2, we get the two dice in part (a); therefore the only solutions have

k pairs of dice as in (a), plus n

2k ordinary dice, for some k 6

in.

558 ANSWERS TO EXERCISES

8.37 The number of coin-toss sequences of length n is F,-‘, for all n > 0,

because of the relation between domino tilings and coin flips. Therefore the

probability that exactly n tosses are needed is

F,-’

/2n, when the coin is fair.

Also

q,,

Fn+‘/2n-‘,

since xkbnFnzn = (Fnzn +

F,-~z~+‘)/(l

zz).

(A systematic solution via generating functions is, of course, also possible.)

8.38

When k faces have been seen, the task of rolling a new one is equivalent

to flipping coins with success probability

= (m

k)/m. Hence the pgf is

nki(,Pkd(l

qkz)

nk:zO(

’

k)z/(m

kz). The mean is ~~~~pk’ =

M-b

H,-l);

the variance is m’(H!’

HE!,)

m(H,

H,-1); and

equation (7.47) provides a closed form for the requested probability, namely

mpnm!{;I,‘}/(

m-L)!. (The problem discussed in this exercise is traditionally

called “coupon collecting!‘)

8.39 E(X) = P(-1); V(X) = P(-2)

P(-l)2;

E(lnX) = -P’(O).

8.40 (a) We have

n(O!{';}p-l!{T}p2

+2!{y}p3

-...),

by (7.49).

Incidentally, the third cumulant is npq(q-p) and the fourth is npq(l-6pq).

Theidentity q+pet = (p+qept)et shows that f,,,(p) =

(-l)“f,(q)+[m=l];

hence we can write

f,,,(p) = g,,,(pq)(q-p)'m

Odd], where

g,,,

is a polynomial

of degree [m/2], whenever m > 1. (b) Let p =

and F(t) = ln(

Set).

Then~,~,~,tm-‘/(m-1~t!=F’(t)=1-1/(et+1),andwecanuseexercise

6.23.

8.41

G(z)

is the pgf for a random variable X that assumes only positive

integer values, then s:

G(z)

dz/z =

tk>,

Pr(X=k)/k = E(X-‘). If X is the

distribution of the number of flips to obtain n + 1 heads, we have G(z) =

(PZ/(l

qz)y+’

(8.5g),

and the integral is

if we substitute w =

pz/(l

qz). When p = q the integrand can be written

(-l)n((l+w)~1-l+w-w2+~~~+(-l)nwn~’),sotheintegralis

(-l)n(ln2-

l+~-~+...+(-l)n/n).

WehaveH~,-Hn=ln2-~n~‘+~n~Z+O(n~4)

(g.28),

and it follows

tha.t

E(X,:,) =

:n-’

$np2

O(np4).

8.42 Let F,(z) and G,(z) be pgf’s for the number of employed evenings, if

the man is initially unemployed or employed, respectively. Let

= 1

and

:= 1

-pf.

Then

Fo(z)

Go(z)

= 1, and

F,,(z) =

PhZGv’

(Z)

+ qhFm

’

(Z)

;

G,,(Z)

PfFn-1 (z)

qfZGn-1

(z).

A ANSWERS TO EXERCISES 559

The solution is given by the super generating function

G(w,z)

G,(z)w”

A(w)/(l

zB(w)) ,

n30

where B(w) =

w(qf-(qf-p,,)w)/(l

-qhw)

and A(w) = (1

-B(w))/(l

-w).

Now tnaO GA(l)w” = cxw/(l

-w)~+

B/(1

-w)

B/(1

(qf

-ph)w)

where

OL=Ph

Ph+Pf

’

Pf(qf

-Ph)

(Ph+Pf)’

’

hence

G;(l)

cxnf

fi(l

(qf-ph)n).

(Similarly

G:(l)

K2n2

+ O(n), so

the variance is O(n).)

8.43

G,(z)

k20

[L].zk/n!

z?/n!,

by (6.11). This is a product of

binomialpgf’s,

nE=,

((k-l

+z)/k),

ere the kth has mean 1 /k and variance

(k-

1)/k’;

hence

Mean

and

Var(G,)

Ht).

8.44 (a) The champion must be undefeated in n rounds, so the answer is

pn.

(b,c) Players

xl,

. . . ,

X~L

must be “seeded” (by chance) in distinct subtour-

naments and they must win all 2k(n

k) of their matches. The 2” leaves of

the tournament tree can be filled in

2n!

ways; to seed it we have 2k!(2”-k)2k

ways to place the top

players, and (2”

2k)! ways to place the others.

Hence the probability is (2p)2k’n-k)/(iE). When k = 1 this simplifies to

(2~~)“~l/(2”

1). (d) Each tournament outcome corresponds to a permuta-

tion of the players: Let

be the champ; let

be the other finalist; let

and

be the players who lost to

and

in the semifinals; let

(ys,

. . . ,ys) be

those who lost respectively to

(y,

, . . . ,y4) in the quarterfinals; etc. (Another

proof shows that the first round has 2n!/2n-1! essentially different outcomes;

the second round has 2np1!/2np2!; and so on.) (e) Let

be the set of

2kp1

potential opponents of x2 in the kth round. The conditional probability that

x2 Wins, given that

belongs to Sk, is

Pr(xl

plays x2)

.pn-’

-p)

Pr(xl

doesn’t play x2) .p”

= pkP’pnP’(l -p) + (1

-pypn.

The chance that

is 2kp1/(2n

1); summing on k gives the answer:

*;,

-$&(pk-‘p+p)

(I-pk-‘)pn)

(‘,“n’“,

’

pnp’

(f) Each of the

2”!

tournament outcomes has a certain probability of occur-

ring, and the probability that

wins is the sum of these probabilities over

all (2”

1 )! tournament outcomes in which xj is champion. Consider inter-

changing xj with xj+l in all those outcomes; this change doesn’t

affect

the

560 ANSWERS TO EXERCISES

probability if xi and

xi+1

never meet, but it multiplies the probability by

p)/p

if they do meet.

8.45 (a) A(z) =

l/(3

22);

B(z)

= zA(z)‘; C(t) = z’A(z)~. The pgf for

sherry when it’s bottled is

ZEAL,

which is

times a negative binomial

distribution with parameters n = 3, p =

(b) Mean(A) = 2, Var(A) = 6;

Mean(B) = 5, Var(B) = 2Var(A) = 12; Mean(C) = 8, Var(C) = 18. The

sherry is nine years old, on the average. The fraction that’s 25 years old is

(,;)

(-21L23

25 = (ii)2223

= 23.

(324

.00137. (c) Let the coefficient of

w” be the pgf for the beginning of year n. Then

A = (1

;w/(l

-w))/(l

{zw);

B = (l+ ;zwA)/(l-

$zw);

C = (1 +

;zwB)/(l

$zw).

Differentiate with respect to z and set z = 1; this makes

C’ =

l/2

3/2

l-w

~ijw!3-(l~~w)2

l-iw’

The average age of bottled sherry n years after the process started is 1 greater

than the coefficient of w”~‘, namely

9-(

f)“(3n2+21n+72)/8.

(This already

exceeds 8 when n = 11.)

8.46

(a)

P(w,z) = 1 +

i(wP(w,z)

zP(w,z)) = (1

i(w

z))~',

hence

mn =

“J

“(“,‘“).

(b) P

w,z) =

i(w"

+zk)P(w,z);

hence

(

,m,n

2k-’ In n

,(,,,,)

(‘“in”-k)).

(C)

tkkPk,n.TI

EL=,

k2km2n(2”;k) = ,YzYo(n

k)2-

k(“zk); this can be

summed using (5.20):

g2mnmk(12n+ll(nn+k)

-(n+l)(nn+:Tk))

= (2n+

l)-(n+1)2-"

(

n+l-2-n-l

2n+2

(

n+l

2n+l

(

22n

(The methods of Chapter 9 show that this is

1 + O(nP’/2).)

8.47 After n irradiations there are n + 2 equally likely receptors. Let the

random variable X, denote the number of diphages present; then

Xn+l

A ANSWERS TO EXERCISES

561

X, + Y,, where

Y,,

= -1 if the (n +

1)st

particle hits a diphage receptor

(conditional probability

2X,/(n

+ 2)) and

Y,,

= +2 otherwise. Hence

EX,,+I

= EX,, + EY, = EX, ~

2EX,/(n+2)

+ 2(1 ~

2EX,/(n+2))

The recurrence

(n+2)EX,+l

(n-4)EX,+2n+4

can be solved if we multiply

both sides by the summation factor (n +

1)“;

or we can guess the answer and

prove it by induction: EX, = (2n + 4)/7 for all n > 4. (Incidentally, there

are always two diphages and one triphage after five steps, regardless of the

configuration after four.)

8.48 (a) The distance between frisbees (measured so as to make it an even

number) is either 0, 2, or 4 units, initially 4. The corresponding generating

functions A, B, C (where, say,

[z”]

C is the probability of distance 4 after n

throws) satisfy

A = $zB, B =

;zB++zC,

C = 1 + ;zB + +zC

It follows that A = z2/(16

202 + 5z2) = t’/F(z), and we have Mean(A) =

Mean(F) = 12, Var(A) =

Var( F) = 100. (A more difficult but more

amusing solution factors A as follows:

PlZ P2Z P2 PlZ

Pl P2Z

-.-=

1-q1z

l-q2z

Pz-Pl

l-412

q2=

’

where

= a2/4 = (3 +

&i/8,

$j2/4

= (3

&)/8,

and

Thus, the game is equivalent to having two biased coins whose

heads probabilities are

and ~2; flip the coins one at a time until they

have both come up heads, and the total number of flips will have the same

distribution as the number of frisbee throws. The mean and variance of the

waiting times for these two coins are respectively 6 F

2fi

and 50 F

22fi,

hence the total mean and variance are 12 and 100 as before.)

(b) Expanding the generating function in partial fractions makes it

possible to sum the probabilities. (Note that

d/(4@)

+ a2/4 = 1, so the

answer can be stated in terms of powers of

4.)

The game will last more than

n steps with probability

5inp11/24pn

( $nf2

@m”p2);

when n is even this is

5"124

nF,+Z.

So the answer is

5504

1ooF,02

.00006.

8.49 (a) If n > 0, PN(O,n) =

i[N=Ol

+ ~PN j(O,n) +

iPN~-i(l,n-1);

(m, 0) is similar;

(0,O)

01.

Hence

gm,n = $=gm l,n+l + ;=%n,n +

&ll+1,n

;

izgg,,

igl,,-1

;

etc.

gk,., =

l-t&,

l,n+l+tg~,,+~g~sl,n~~l;

sh,, =

~+$&,+-&&

+?tc.

By induction on m, we have

gi,,

= (2m +

l)gh,,,+,,

2m2 for all

m,n

3 0.

562 ANSWERS TO EXERCISES

And since

gk,O

91,

we must have gk,,, =

m+n+2mn.

is satisfied when mn > 0, because

sin(2mf

l)e

(

sin(2m-

l)tl

sin(2mt

l)G

sin(2m+3)8

’

this is a consequence of the identity sinjx

y) i- sinjx + y) = 2sin x

cosy.

all that remains is to check

t.he

boundary conditions.

8.50 (a) Using the hint, we get

3(1

42;

(‘f)

(;z)Yl

-Z)2

3(1

-z)2F

(‘f)

(i)kF

(ktjp3)zj+k;

now look at the coefficient of z3+‘. (b) H(z)

+ &z +

c~+~z~+~.

J(1

-2)(9-z).

ne can show that

(z-3+r)(z-3-r)

~42,

and hence that

(r/(1

-2)+2)’

(13-5z+4r)/(l-z)

(9-H(z))/(l -H(z)).

(d) Evaluating the first derivative at z = 1 shows that Mean(H) =

The

second derivative diverges at z =

so the variance is infinite.

8.51 (a) Let H,(z) be the pgf for your holdings after n rounds of play, with

Ho(z) = z. The distribution for n rounds is

H,,+liz)

H,(W)

so the result is true by induction (using the amazing identity of the preceding

problem). (b)

g,,

=H,(O)-H,

l(O)

=4/n(n$-l)(n+2)

=4(n-l)A.

The

mean is 2, and the variance is infinite. (c) The expected number of tickets you

buy on the nth round is Mean(H,,) =

by exercise 15. So the total expected

number of tickets is infinite. (Thus, you almost surely lose eventually, and you

expect to lose after the second game, yet you also expect to buy an infinite

number of tickets.) (d) Now the pgf after n games is

H,(z)‘,

and the method

of part (b) yields a mean of 16

4x2

2.8. (The sum

t,,,

l/k2

7rL/6

shows up here.)

8.52 If w and w’ are events with Pr(w) >

Pr(w’),

then a sequence of

n independent experiments

.will

encounter

more often than w’, with high

probability, because w will occur very nearly

nPr(w)

times. Consequently,

as n

co,

the probability approaches 1 that the median or mode of the

A ANSWERS TO EXERCISES 563

values of X in a sequence of independent trials will be a median or mode of

the random variable X.

8.53 We can disprove the statement, even in the special case that each

variableisOor1.

LetpO=Pr(X=Y=Z=O),pl=Pr(X=Y=Z=O),...,

p7=Pr(X=Y=Z=O),whereX=l-X.

Thenpo+pl+...+p7=1,and

the variables are independent in pairs if and only if we have

(p4+p5+p6+p7)(pL+p3+p6+p7)

p6+p7,

(p4

+ p6 -+

p7)(pl

+ p3 + p5 +

p7)

p7,

(p2

+ p3

-t

p7)tpl

+ p3 +

p7)

= p3 +

p7.

But

WX+Y=Z=O)

Pr(X+Y=O)Pr(Z=O)

(pO

+p,)(pc

+ p4 +

p61.

One solution is

PO = P3

P6 =

l/4;

p1 =

= p4

This is equivalent to flipping two fair coins and letting X = (the first coin

is heads), Y = (the second coin is heads), Z = (the coins differ). Another

example, with all probabilities nonzero, is

PO = 4/64, PI = ~2

~4 =

5/64,

p6 = 10/64,

= 15/64.

For this reason we say that n variables XI, , X, are independent if

Pr(X1

=x1

and...and

Xn=x,)

= Pr(X,

=xl)...Pr(X,

= x,);

pairwise independence isn’t enough to guarantee this.

8.54 (See exercise 27 for notation.) We have

E(t:)

= nll4

+n(n-1)~:;

E(LzLfI

np4

+2n(n-l)u3pl

+n(n-1)~:

+n(n-l)(n-2)p2&;

E(xy) = np4

+4n(n-l)p~u1

+3n(n-1)~:

6n(n-l)(n-2)u2p:

n(n-l)(n-2)(np3)pT

;

it follows that V(\iX) = K4/n +

2K:/(n

~ 1).

8.55 There are A q = & .52! permutations with X = Y, and B =

.52!

permutations with X # Y. After the stated procedure, each permutation

with X = Y occurs with probability

A/((

gp)A),

because we return

to step

with probability

$p.

Similarly, each permutation with X # Y

occurs with probability

g(l

p)/((l ~

sp)B).

Choosing p =

makes

= x and Y = 9) =

for all x and

(We could therefore make two flips

of a fair coin and go back to

if both come up heads.)

564 ANSWERS TO EXERCISES

8.56 If m is even, the frisbees always stay an odd distance apart and the

game lasts forever. If m =

i:l.

+ 1, the relevant generating functions are

(The coefficient

[z”]

is the probability that the distance between frisbees

is 2k after n throws.) Taking a clue from the similar equations in exercise 49,

we set

= 1 /cos’

and

X sin28, where X is to be determined. It follows

by induction (not using the equation for Al) that

= X sin2kO. Therefore

we want to choose X such that

(

1-p

4cos28

X sin;!10 = 1 +

sin(21-

218.

It turns out that X = 2

cos’

O/sin

cos(21+ 1

)O,

hence

cos e

cos

’

The denominator vanishes when

is an odd multiple of

n/(2m);

thus 1

-qkz

a root of the denominator for 1 6 k 6

and the stated product representation

must hold. To find the mean and variance we can write

Trigonometry wins

again.

there a

L.04

. .

)/(I

$2@

&m4@4

. . . )

connection with

pitching pennies

along the angles of

the m-gon?

= 1 +

i(m2

1)02

+ &(5m4

-6m2+

1)04

+...

= 1

+~(m2-l)(tanB)2+~(5m4-14m2+9)(tan8)4+~~~

= 1 +

G:,(l)(tan8)2

iGK(1)(tan8)4

+...

because tan28 =

Z-

1 and tan8

=O+

i03

+....

So we have

Mean

i(m2-1)

andVar(G,)

im2(m2-1).

(Note that thisimplies theidentities

(m-1

l/2

'mjf'2(l/sin

"",r,,')")';

k=l

m2(m2

Ii/2

cot

(2k-

1)~

(2k-

1)n

sin

’

k=l

The third cumulant of this distribution is &m2(m2

l)(4m2

1); but the

pattern of nice cumulant factorizations stops there. There’s a much simpler

A ANSWERS TO EXERCISES 565

way to derive the mean: We have

= z(Ar

. . +

AL)

hence when

= 1 we have

+ Al. Since

= 1 when

= 1, an

easy induction shows that

= 4k.)

8.57 We have A:A 3

2’

’

and B:B < 2l

’

+ 2l

and B:A 3 2’

hence

13:B

B:A 3 A:A

A:B is possible only if A:B >

21p3.

This means that

~3,

~4,

‘~5,

. , rr

= rr. But then A:A

2’

’

2’

..I

A:B

2’

, B:A

2’

’

2’

. . , and B:B

2’

-’

2’

+ . . .

;

hence B:B

B:A is less than A:A

A:B after all. (Sharper results have

been obtained by Guibas and Odlyzko

[138],

who show that Bill’s chances are

always maximized with one of the two patterns

H-r1

. . .

or Trl

8.58 According to

r(8.82),

we want B:B

B:A > A:A

A:B. One solution is

A = TTHH, B = HHH.

8.59 (a) Two cases arise depending on whether

# h, or

= h,:

m-l

m-2+w+z

k-1

G(w,z)

---(

(

m-l+2

nmk

(

rn-cwzJk

lwZ~m.:;S

k-lZ.z

(b) We can either argue algebraically, taking partial derivatives of G (w, z)

with respect to w and

and setting w =

= 1; or we can argue com-

binatorially: Whatever the values of

hl,

. . . , h,

-1,

the expected value of

P(hl , . , h,

h,;

n) is the same (averaged over h,), because the hash se-

quence

(hr

, . . . , h, 1

)

determines a sequence of list sizes (nl , n2,. . , n,) such

that the stated expected value is

((nr+l)

(nz+l)

...

(n,+l))/m

1 +

m)/m.

Therefore the random variable EP(

, . . . , h,,; n) is indepen-

dent of (hl , . , h, I), hence independent of P(

, . . ,

h,;

k).

8.60 If 1 6 k <

n, the previous exercise shows that the coefficient of

sksr

in the variance of the average is zero. Therefore we need only consider

the coefficient of si, which is

Pih,,...,h,;k)2

-’

1Sh1

,...I

h,,Sm

l<h,

,...I

h,,<m

the variance of ((m

1 +

z)/m)

k~’

z; and this is (k

l)(m

1)/m’

as in

exercise 30.

8.61 The pgf D,(z) satisfies the recurrence

Do(z) =

b(z)

= z2Dn

(2)

2(1

z3)Dk

(z)/(n + 1))

for n > 0.

566 ANSWERS TO EXERCISES

We can now derive the recurrence

D:(l)

(n-

ll)D,!P,(l)/(n+

1) +

(8n-2)/7,

which has the solution

(n+2)

(26n+ 15) for all n 3 11 (regardless of initial

conditions). Hence the variance comes to

(n + 2)(212n + 123) for n 3

11.

8.62 (Another question asks if a given sequence of purported cumulants

comes from any distribution whatever; for example, ~2 must be nonnegative,

and ~4 +

3~:

= E((X

~1~)

must be at least (E((X

FL)‘))’ = K:, etc.

A necessary and sufficient condition for this other problem was found by

Hamburger

[6],

[144].)

8.63 (Another question asks if there is a simple rule to tell whether H or T

is preferable.) Conway conjectures that no such ties exist, and moreover that

there is only one cycle in the directed graph on

2’

vertices that has an arc

from each sequence to its “best beater!’

9.1 True if the functions are all positive. But otherwise we might have,

say, fl (n) =

+ n2, fz(n) =

-n3,

(n) =

+ n, g2(n) =

-n4.

9.2 (a) We have nlnn 4

c”

4 (Inn)“, since (lnn)2 +

nlnc

4 nlnlnn.

(b)

nlnlnlnn

4 (Inn)! + nlnlnn. (c) Take logarithms to show that (n!)! wins.

‘$,,,

2lnn

,2lnl$.

,,-nln$winsbecause@‘=@+l

<e.

9.3 Replacing kn by 0 (n) requires a different C for each k; but each 0

stands for a single C. In fact, the context of this 0 requires it to stand for

a set of functions of two variables k and n. It would be correct to write

,Tc=, kn =

EL=,

O(n2) = O(n3).

9.4

For example, limn+03

0(1/n)

= 0. On the left,

0(1/n)

is the set of all

functions f(n) such that there are constants C and

with If(n)1 < C/n for

all n 3

no.

The limit of all functions in that set is 0, so the left-hand side is

the singleton set {O}. On the right, there are no variables; 0 represents {0}, the

(singleton) set of all

“f

unctions of no variables, whose value is zero!’ (Can you

see the inherent logic here? If not, come back to it next year; you probably

can still manipulate O-notation even if you can’t shape your intuitions into

rigorous formalisms.)

9.5 Let f(n) =

and g(n) = 1; then n is in the left set but not in the

right, so the statement is

fa.lse.

9.6

nlnn+yn+O(filnn).

9.7 (1

-em’/n)P’

=nBo-B1

+B2n~~‘/2!+~.~=n+~+O(n

‘).

9.8 For example, let f(n) =

[n/2]!’

+n,

g(n) =

([n/2]

l)! [n/2]!

+n.

These functions, incidentally, satisfy f(n) = O(ng(n)) and g(n) =

O(nf(n));

more extreme examples are clearly possible.

A ANSWERS TO EXERCISES 567

9.9 (For completeness, we assume that there is a side condition n

00,

so that two constants are implied by each 0.) Every function on the left has

the form a(n) + b(n), where there exist constants

B, no, C such that

la(n)/ 6 Blf(n)[ for n 3 mc and [b(n)1 6 Clg(n)l for n 3 no. Therefore the

left-handfunctionisatmostmax(B,C)(lf(n)l+Ig(n)l),forn3max(~,no),

so it is a member of the right side.

9.10 If g(x) belongs to the left, so that g(x) = cosy for some y, where

Iy/

< Clxl for some C, then 0 6 1

g(x) = 2sin2(y/2) <

$y2

6 iC2x2; hence

the set on the left is contained in the set on the right, and the formula is true.

9.11

The proposition is true. For if, say,

1x1

/yI,

we have (x +

Y)~

6 4y2.

Thus

(x+Y)~

0(x2)

+O(y’).

Thus

O(x+y)’

O((x+y)‘)

0(0(x2)

O(y2)) =

0(0(x2))

-t O(O(y2)) =

0(x2)

+ O(y2).

9.12 1 +2/n + O(nP2) = (1 +

2/n)(l

+ O(nP2)/(1 +2/n)) by (g.26), and

l/(1

+2/n) = O(1); now use (9.26).

9.13 n”(1 + 2nP’ + O(nP2))” =

nnexp(n(2n-’

+ O(nP2))) = e2nn +

O(n”-‘).

9.14

It is

nn+Pexp((n+

@)(ol/n- ta2/n2

+O(ne3)))

9.15

In (n2n) =

3nln3

the answer

is‘

nn+tln3-ln2n+

(+f)n-’

+O(nP3),

=(I

5n-l

82jnp2

o(n-3)).

9.16

is any integer in the range a 6

< b we have

B(x)f(l+x)

dx =

B(x)f(l+x)

dx-

l/2

B(l

-x)f(l+x)dx

B(x)(f(l+x)

-f(l+

-x))

dx.

l/2

Since

+ x >

+ 1

x when x 3

this integral is positive when f(x) is

nondecreasing.

9.17

L>O

B,(i)z."'/m!

= ~e~'~/(e~-l) =

z/(eZ/2-1)-z/(e"-1)

9.18

The text’s derivation for the case

= 1 generalizes to give

2(2n+1/2)a

bk(n) = --e

-k’a/n

(27rn)"/2

ck(n) = 22nan

-(l+cx)/2+3ykb./n.

the answer is 22na(~n)i’~a1’20L~1’2(1 + O(n-1/2+36)).

568 ANSWERS TO EXERCISES

9.19

Hlo = 2.928968254

2.928968256;

lo!

3628800

3628712.4;

B,,.,

0.075757576

0.075757494;

10) = 4

10.0017845;

e".'

= 1.10517092

1.10517083;ln1.1 = 0.0953102

0.0953083; 1.1111111

1.1111000~

l.l@.'

1.00957658

1.00957643. (The approximation to n(n) gives more significant

figures when

is larger; for example,

rc(

09)

= 50847534

50840742.)

9.20 (a) Yes; the left side is o(n) while the right side is equivalent to O(n).

(b) Yes; the left side is e. eoi’/ni. (c) No; the left side is about

J;;

times the

bound on the right.

9.21 WehaveP,=m=n(lnm-1

-l/lnm+O(l/logn)2),

where

lnm = lnn+lnlnm-

l/lnn+lnlnn/(lnn)2

+O(l/logn)2;

lnlnn

(lnlnn)’

lnlnm =

1nlnn-t

-In

lnlnn

2(lnn)2

(lnn)2

O(l/logn)‘.

It follows that

= n

(

lnn+lnlnn-1

lnlnn-2

t(lnlnn)’

31nlnn

---

(lnn)2

O(l/logn)’

)

(A slightly better approximation replaces this 0(

l/logn)’

by the quantity

-5/(lnn)’

+ O(loglogn/logn)3; then we estimate

P~OOOOOO

15483612.4.)

9.22 Replace

O(nzk)

by --&npLk +

O(n

4k)

in the expansion of

H,r;

this

replaces

O(t3(n2))

-h.E3(n2)

+ O(E:3(n4)) in (9.53). We have

,X3(n)

= ii-i-

’

&n,F2

+ O(np3),

hence the term

O(n2)

($1.54)

can be replaced by

-gnp2

+ O(n 3).

g.23 nhn =

toskcn

hk/(n~-k)

+ZcH,/(n+

l)(n+2).

Choose c =

enL/6

tkaogk

that

tka0

0 and h, = O(log n)/n3. The expansion of

OSk<n

hk/(n

k) as in (9.60) now yields nh, =

ZcH,/(n

l)(n

+ 2) +

O(n

m2),

hence

en~/6

n+2lnn+O(l)

(

.n3

9.24

(a)

,&o(f(k)

< co and if

f(n

O(f(n)) when 0 6 k < n/2,

we have

akhk

O(f(k))O(f(n))

O(f(n))O(f(n

k)) ,

k=O k=O

k=n/2

A ANSWERS TO EXERCISES 569

which

2O(f(n)

tkzO

If(k)/),

so this case is proved. (b) But in this case if

a - b, = aPn, the convolution (n + 1 )aPn is not 0(

“).

9.25

s,/(3t)

~;4Lq2n+l)F

may restrict the range of summation

to 0 < k 6 (logn)‘, say. In this range nk =

nk(l

(i)/n

+ O(k4/n2)) and

(2n +

l)k

= (2n)k(l +

(“;‘)/2n+

O(k4/n2)), so the summand is

Hence the sum over k is 2 -4/n + 0( 1 /n2). Stirling’s approximation can now

be applied to

(y)

= (3n)!/(2n)!n!, proving (9.2).

9.26 The minimum occurs at a term

Blm/(2m)

(2m-

1 )n2”-’ where 2m

2rrn + 3, and this term is approximately equal to 1

/(rceZnnfi

The absolute

error in Inn! is therefore too large to determine n! exactly by rounding to an

integer, when n is greater than about

e2n+‘.

9.27 We may assume that a #

n”+l

C,+

k=l

a+1

Let f(x) = x”; the answer is

na-2k+l

0~~”

-2m

1).

(The constant

turns out to be

<(-a),

which is in fact defined by this

formula when a > -1.)

9.28 Take f(x) = xlnx in Euler’s summation formula to get

nnL:2+n/:+1/12e~n~i4(1

qn-2))

where A

1.282427 is “Glaisher’s constant!’

9.29 Let f(x) =

xP1

lnx. Then fiZmi (x) > 0 for all large x, and we can write

Ink

y+lnS+z+Bn+,

0<8,<1,

k=l

where S

0.929772 is constant. Taking exponentials gives

(In general if f(x) =

lnx, Euler’s summation formula applies as in exer-

cise 27, and the resulting constant is -<‘(-a) if a # -1. Thus, the theory of

the zeta function gives a closed form for Glaisher’s constant in the previous

exercise. We have

1nS

= yi in the notation of answer 9.57.)

570 ANSWERS TO EXERCIS:ES

9.30 Let g(x) =

xLePxL

and. f(x) =

g(x/fi).

Then n “’

,Yk>O

k’ePkz”’ is

f(x)

%‘kP”(q

(-1

)-I

Oc’

h&4)

k=l

,fl"'(x)

l/2

g(x)

E!Lnlk~l)i2gik-l1(0) + 0(~-m/2).

k=,

Since g(x) = x1

x2+‘/l ! + x4 ‘l/2!

x6+‘/3!

. , the derivatives g imi (x) obey

a simple pattern, and the answer is

1,it+l)/2

’

(

Bt+l

b+3np’

Bt+6

2 2

(1+1)!0!

(l+3)!1!

(1+5)!2!

+Obp3)

9.31 The somewhat surprising identity

l/(cmmmk

cm)

l/(~"'+~

cm)

1 /cm makes the terms for 0 < k 6 2m sum to (m + +)/cm. The remaining

terms are

1 1

C2m+l

C2m

C3m+2

C3m

+...

)

and this series can be truncated at any desired point, with an error not ex-

ceeding

the first omitted term.

9.32

H:)

x2/6

l/n + O(nP2) by Euler’s summation formula, since we

know the constant; and

is given by (9.89). So the answer is

The world’s top

three constants,

ney+nL’6

1 -

in-’

O(n-‘))

(

(e,

y),

all

appear

in this answer.

9.33

Wehavenk/n’=

l-k.(k-l)nP’+~k2(k--l)2n~2+0(k6nP3);

dividing

by k! and summing over k 3 0 yields e

en-’

en-

+ 0 ( nP3 ) .

9.34 A =

ey;

B = 0; C =

-.ie’;

D =

ieY(l

-y);

E = :eY; F =

&eY(3v+l).

9.35 Since l/k(lnk+

O(l])

l/kink+

O(l/k(logk)2),

the given sum

is Et==, 1

/kink

+ 0( 1).

The remaining sum is In Inn + 0( 1) by Euler's

summation formula.

9.36 This works out beautifully with Euler’s summation formula:

+L--

-2x

+ x2 2

+ x2

+?(n2+x2)2

+ O(nm5)

A ANSWERS TO EXERCISES 571

Hence

a7m-l

-- inP2

An3

+ O(nP5).

9.37 This is

k,q>l

n2-

= ,2

The remaining sum is like (9.55) but without the factor u(q). The same

method works here as it did there, but we get L(2) in place of

l/<(2),

SO the

answer comes to (1

g)nZ

+ O(nlogn).

9.38 Replace k by n

k and let ok(n) = (n

k)nPk(f;). Then In ok(n) =

nlnn

Ink!

k +

O(kn’),

and we can use tail-exchange with bk(n) =

nnePk/k!, ck(n) = kbk(n)/n,

= {k 1 k < lnu}, to get

ok(n) =

nne’/e(l +

O(n’)).

9.39 Tail-exchange with bk(n) = (Inn

k/n

ik2/n2)(lnn)k/k!,

ck(n) =

(In n) k+3/k!,

={k 1 0 < k <

10lnn).

When k x 1Olnn we have

k! x

fi(lO/e)k(lnn)k,

so the kth term is

O(n-

101n(lO/e)

logn). The answer

is nlnn-lnn-

t(lnn)(l

+lnn)/n+O(n~2(logn)3).

9.40 Combining terms two by two, we find that

H&-(H2k-&)m

= EHykP’

plus terms whose sum over all k > 1 is 0 (1). Suppose n is even. Euler’s

summation formula implies that

hence the sum is

H,”

+ 0 (1). In general the answer is

9.41 Let

CX=

$/L$ = -@-2. We have

(In eYn)m

+0(l)

-l)nH,m -t O(1).

ClnFk

~(h~k-h&+h(l

-ak))

k=l

n(n

+ 1)

In@-5ln5+tln(l

-ak)-xln(l

-elk).

k21

k>n

The latter sum is tIk>,,

O(K~) =

O(~L~).

Hence the answer is

@+1/25-Wc

o&n’”

31/+-n/Z) ,

where

C = (1

-a)(1

-~~)(l

-K~)...

1.226742.

572 ANSWERS TO EXERCISES

9.42 The hint follows since

(,“,)/(z)

Let

m =

lcxn]

Then

(

1+i~+(&)2+...)

(;)S.

ksa,,

(;)

(:)0(l),

an i remains to estimate

(z).

By Stirling’s ap-

proximation we have In

(z)

-i

nn-(an-e)ln(K-e/n)-((l--0()n+c)

ln(l-cx+c/n)+0(1)=-~lnn-omlna-(1-ol)nIn(l-cx)+0(1).

9.43 The denominator has factors of the form

w, where w is a complex

root of unity. Only the factor

1 occurs with multiplicity 5. Therefore

by (7.31), only one of the roots has a coefficient

n(n4),

and the coefficient is

c =5/(5!~1~5~10~25~50)=1/1500000.

9.44 Stirling’s approximation says that

ln(xP”x!/(x-a)!)

has an asymptotic

series

-a-(x+i-a)ln(l-a/x)-&(x

‘-(x-o())‘)

&(x

cc)

“)

-’

in which each coefficient of

xm~k

is a polynomial in

(x.

Hence x

“x!/(x

CX)!

Co(R)

+c1(a)x

’

...

+ c,(tx)xpn +

0(x-”

‘) as x

03, where c,,(a) is a

polynomial in

01.

We know that

(

LX)

[,*,I

(-1)" whenever

is an integer,

and

LA1

is a polynomial in

of degree 2n; hence

(

CX)

= [

&*,,I

(-1)” for

all real

01.

In other words, the asymptotic formulas

generalize equations (6.13) and

(6.11),

which hold in the all-integer case.

9.45 Let the partial quotients of LX be

(a,,

al,.

. . ), and let

cc,,,

be the con-

tinued fraction

l/(a,

CX,,~,)

for m 3 1. Then D(cx,n) = D(cxl,n) <

D(olr, LarnJ) +

D(tx3,

LcxzlcxlnJj) + al +

...

D(Lx,+I,

~~m~...~~,n~...~~)+a~+~..+a,+3m<oll...cx,n+al+...+a,+3m,

for all m. Divide by n and let n

co;

the limit is less than

011

. . .

CX,

for

all m.

Finally we have

011

.a,

9.46 For

convenien.ce

we write just m instead of m(n). By Stirling’s ap-

A ANSWERS TO EXERCISES 573

proximation, the maximum value of

k:/k!

occurs when k

n/inn,

we replace k by m + k and find that

Cm+

kin

In 27rm

(m-t

k)!

nlnm-mlnmfm-P

(m+n)k2

2m2

+ O(k3m ‘logn)

Actually we want to replace k by

[ml

+ k; this adds a further 0 (km

’

log n).

The tail-exchange method with

lkl

< m’/2+E now allows us to sum on k,

A truly

Be/l-shaped

giving a fairly sharp asymptotic estimate

summand.

= --

The requested formula follows, with relative error 0 (log log n/log n).

9.47

Letlog,n=l+El,whereO$8<1.

Thefloorsumisl(n+l)+l-

(ml+’

l)/(m

1):.

the ceiling sum is

+ 1)n

(ml+’

l)/(m

1); the

exact sum is (1+

0)n

n/in

m + O(log n). Ignoring terms that are o(n), the

difference between ceiling and exact is ( 1

f (0)) n, and the difference between

exact and floor is f(O)n, where

f(e)

J!&Y+e----.

lnm

This function has m,aximum value f (0) = f (1) = m/( m

/In

m, and its

minimum value is

lnlnm/lnm

+ 1

(ln(m

l))/ln m. The ceiling value is

closer when n is nearly a power of

but the floor value is closer when

lies

somewhere between 0 and 1.

9.48 Let

+ bk, where

counts digits to the left of the decimal

point. Then

= 1 +

Llog

Hk] = log log k + 0( 1 ), where ‘log’ denotes

loglo.

To estimate bk, let us look at the number of decimal places necessary to

distinguish y from nearby numbers y --

and y + E’: Let 6 = 10

be the

574 ANSWERS TO EXERCISIES

length of the interval of numbers that round to 0. We have

-01

6 id; also

y-e <

Q--i6

andy+c’

> Q-t-:8. Therefore e+c’ > 6. Andif 6 < min(e, E’),

the rounding does distinguish ij from both y

and y + 6’. Hence

10Ph”

l/(k-l)+l/kand

10IPbk

3 l/k; we have

= log k+O(l). Finally, therefore,

Et=,

(logk+loglogk+O(l)), which is nlogn+nloglogn+O(n)

by Euler’s summation formula.

9.49 We have

lnn+y+

in-’

&nP2 = f(n), where f(x) is increasing

for all x > 0; hence if n 3 ea

we have

3 f(e”-Y) >

Also

H,-,

Inn + y

in--’ =

g(n), where g(x) is increasing for all x > 0; hence if

n 6 eaPy we have

H,-l

$ g(e”--Y)

01.

Therefore

H,-r

implies

that

eaPv+l

>n>ea+Y-l.

(Sharper results have been obtained by Boas

and Wrench [27].)

9.50 (a) The expected return is ,YlsksN

k/(k’HE’)

= HN/H~‘, and we

want the asymptotic value to

O(N-’

1nN

+y+O(N-‘)

6lnlO

6y 361n10 n

n2/6-N-l+O(N-2)

~n+~~+~~+o(lo-n)*

The coefficient (6 In 1

O)/n2

= 1.3998 says that we expect about 40% profit.

(b) The probability o:f profit is x,,<kGN

l/(k2Hc’)

= 1

Hf’/HE’,

and since

Hf)

-n-l

in-’

+ O(nm3) this is

n-’

in2

+O(nP3)

--n

n2/6+

O(N-1) = 7crn

2+0(nP3),

actually decreasing with n. (The expected value in (a) is high because it

includes payoffs so huge that the entire world’s economy would be affected if

they ever had to be made.)

9.51 Strictly speaking, this is false, since the function represented by O(xP2)

might not be integrable. (It might be

‘[x

S]/x”,

where S is not a measurable

set.) But if we stipulate that f(x) is an integrable function such that f(x) =

(As opposed to an

O(xm2) as x

00, then

IJ,“f(x)

dx( <

j,“lf(x)I

dx <

j,”

CxP2 dx =

Cn’.

execrable function.)

9.52 In fact, the stack of n’s can be replaced by any function f(n) that

approaches infinity, however fast. Define the sequence (TQ,

ml,

. . . ) by

setting

rnc

= 0 and letting mk be the least integer > mk-1 such that

3 f(k+

1)‘.

Now let A(z) =

tk>,

(z/k)mk.

This power series converges for all z, because

the terms for k >

Iz/

are bounded by a geometric series. Also A(n + 1) 3

((n+

l)/n)“‘n 3

f(n+l)‘,

hence

lim,,,f(n)/A(n)

=O.

A ANSWERS TO EXERCISES 575

9.53 By induction, the 0 term is (m

l)!--’

s,”

tmP’f(“‘)(x

t) dt. Since

f(ln+‘) has the opposite sign to f

cm),

the absolute value of this integral is

bounded by If(“‘(O) 1

J,”

tm-’ dt; so the error is bounded by the absolute value

of the first discarded term.

9.54 Let g(x) =~f(x)/xrx. Then g’(x)

-oLg(x)/x as x

00. By the mean

Sounds like a nasty

value theorem, g(x

g(x +

= -g’(y)

ag(y)/y

for some y between

theorem.

and x +

Now g(y) = g(x)(l +0(1/x)), so g(x

g(x +

ag(x)/x

af(x)/xlta.

Therefore

f(k)

k3n k’+”

(J(t(g&-

g(k+

iI))

o(g(n-

:I).

k3n

9.55

The estimate of (n + k +

ln(l + k/n) + (n

k +

ln(1

k/n) is

extended to k2/n + k4/6n3 + O(nP3/2+5E), so we apparently want to have an

extra factor

ePk4/6n3

in bk(n), and ck(n) = 22nn-2+5eePk*/n. But it turns

out to be better to leave bk(n) untouched and to let

ck(n)

22nTL-2+5ce-kZ/n +

22nn-5+5~,&-kz/~,

thereby replacing e

-1c4/6n3

by 1

0 (

k4/n3

) .

The sum

k2/n

is 0 (

n512

) ,

as shown in exercise 30.

9.56 If k <

n’/‘+’

we have ln(nk/nk) =

-gk’/n

ik/n

ik3/n2 +

0 (n-

1+4E)

by Stirling’s app roximation, hence

nk/nk = ePkzi2n(l +

k/2n

$k3/(2n)2 +

O(nP”4’))

Summing with the identity in exercise 30, and remembering to omit the term

for k = 0, gives -1 + 01~ +

O:‘,’

$G:“,’

+ O(nP1/2+4’) =

O(n-

1/2+4e)

9.57 Using the

hini;,

the given sum becomes J,”

ueCU<(

1 +

u/inn)

du. The

zeta function can be defined by the series

<(l +

C’

(-l)“r,z’“/m!

Ill>0

where

= y and y,,, is the Stieltjes constant

Hence

the given sum is

576 ANSWERS TO EXERCISES

9.58

Let 0

and

f(z) =

e2xiro/(

eZnir

1).

We have

when xmod 1 = 4;

when

lyl

Therefore

/f(z)1

is bounded on the contour, and the integral is O(Mlmm).

The residue of

2nif(z)/zm

= k # 0 is eznike/km; the residue at

= 0 is

the coefficient of

2-l

e2niz0

2rriz

Zm+l (Bo +

. .

27riz

&,(Wi

+W+

+-.)

namely

(2ti)“‘B,(O)/m!.

Therefore the sum of residues inside the contour is

m,B,(B)

(27ri)m

enim/2

COS

(2nk6 --

nm/2)

kz=l

This equals the contour integral O(Mlpm), so it approaches zero as M -+ 00.

9.59 If

F(x)

is sufficiently well behaved, we have the general identity

F(k + t) =

G(2rm.)eZRint ,

where G(y) = ST,”

eciyXF(x)

dx. (This is “Poisson’s summation formula:

which can be found in standard texts such as Henrici (151, Theorem 10.6e].)

9.60 The stated formula is equivalent to

5 21

___-

+ 1024n3

32768n4

O(C5)

by exercise 5.22. Hence the result follows from exercises 6.64 and 9.44.

9.61 The idea is to make

“almost” rational. Let ok =

22zk

be the kth

partial quotient of

01,

and let n = ;a,,,+, qm, where

K(al,.

. . , a,) and

m is even. Then 0

{q,,,K}

l/Q(al,...,a,+.,)

< 1/(2n), and if we take

v =

a,,,+1

/(4n) we get a discrepancy 3

:a,+,

. If this were less than

n’-’

would have

%+1

WlAy),

but in fact a,+1 >

42,"

A ANSWERS TO EXERCISES 577

“The paradox

is now fully es-

tablished that

the utmost

abstractions are the

true weapons with

which to control

our thought of

concrete fact.”

-A. N.

White-

head [304]

9.62 See Canfield

[43];